Solids by kOk3xc8

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									       New Chapter 13, Solids I: Theory                                                13.1



                                  Chapter 13
                           Solids I: Theory

       13.1 Introduction
       13.2 Bonding of Solids
       13.3 Crystals and Noncrystals
       13.4 Energy Levels of Electrons in a Solid; Bands
       13.5 Conductors and Insulators
       13.6 The Drude Model of Conductivity
       13.7 Electron Collisions in Metals
       13.8 The Fermi Speed
       13.9 Degeneracy Pressure
       13.10 White Dwarfs, Neutron Stars, and Black Holes
       13.11 Classical and Quantum Gases
       13.12 Bose-Einstein Condensation
                 Problems for Chapter 13
       
           Advanced Topics which may be omitted without serious loss of continuity




13.1 Introduction
       In Chapter 12 we saw how atoms can bond together to form molecules. In this
chapter we study the aggregation of atoms on a larger scale, to form solids. There are
many close parallels between solid-state and molecular physics. Several of the bonding
mechanisms in solids are exactly the same as in molecules; for example, many solids are
covalently or ionically bonded. There are, however, many differences between solid-state
and molecular physics. For example, the bonding of metallic solids has no direct parallel
in molecules. Also, certain solids, instead of being simple aggregates of atoms, are better
viewed as being built in two stages, with atoms bound into molecules and then molecules
bound into a solid. (A good example is ice, which is best understood as an aggregate of
water molecules.) Another new feature of the solid state is the possibility that atoms (or
molecules) can bond together into a symmetric, repetitive crystalline array; the study of
the possible crystal structures is an important part of solid-state physics with no exact
counterpart in molecular physics.


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        New Chapter 13, Solids I: Theory                                                             13.2

         Perhaps the most striking thing about solids is that they were the first example of
macroscopic systems that were understood with the help of quantum mechanics.*
Numerous commonplace properties of solids — specific heat, conductivity, transparency,
magnetic susceptibility, compressibility, and many more — had defied classical
explanation, but could be explained and calculated using quantum mechanics. It is this
ability to explain — and hence control — the electronic and mechanical properties of
solids that has led to a vast array of technological applications which have transformed
civilization in the last century. Probably the most important technological advance in
solid-state physics was the invention of the transistor, which led to miniturized
electronics, personal computers, and the digital information age we now live in. The
transistor and several other applications of solid state physics are examined in the next
chapter. In this chapter, we concentrate on the theory underlying these applications.
         The theory of solids necessarily involves approximations. In principle, any
property of any solid can be predicted using Schrödinger's equation, but in practice this
strategy is fruitless. In the case of a macroscopic solid with some 1023 electrons and
nuclei, Schrödinger's equation is exceedingly complex and impossible to solve exactly.
(Even in the case of simple molecules with just a few atoms, Schrödinger's equation
cannot be solved exactly.) The art of solid state theory is to model the system under study
by simply ignoring most of the real-world complexities and concentrate on just a few
essential features. For instance, in computing the electronic properties of metal, one often
assumes that the conduction electons behave as free electrons in a box — thus the theorist
completely ignores the interactions of the electrons with the nuclei and with each other. A
theoretical model must be simple enough to allow calculations, but not so simple as to
erase the properties the theorist wishes to explain. The check that the model is correct,
that its approximations are valid, is always experiment.
        We begin this chapter, in Sections 13.2 – 13.5, with a broad, qualitative overview
of the physics of solids. In Section 13.2 we describe the principal mechanisms by which
atoms or molecules bond to one another to form solids. In Section 13.3 we describe some
of the regular crystalline structures in which many solids can form, and we contrast these
with the irregular, noncrystalline structure of the amorphous solids and the still more
irregular composition of liquids and gases. We next take up what is perhaps the single
most important achievement of solid-state physics, the theory of electrical conductivity.
This theory depends on an understanding of the electron energy levels in a solid. In
Section 13.4 we discuss these levels and describe how they fall into bands of allowed
energies. Within these bands the energy levels are distributed almost continuously, but
between them are gaps where there are no allowed levels. In Section 13.5 we show how
the different possible arrangements of these bands explain why some solids are
conductors, some are insulators, and some — the so-called semiconductors — are in
between.
      Following our introductory survey of the field, we take a closer look at the use of
quantum mechanics in explaining properties of solids, with an emphasis on electronic

        *
          The applications of quantum theory to solids go back to the 1907 paper of Einstein, in which he
used quantization to explain the specific heats of solids. See Section 15.10.


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       New Chapter 13, Solids I: Theory                                                13.3

properties. In Sections 13.6 – 13.8, we examine a relatively simple but very useful model
of electrical conduction called the Drude model. We will see that, at almost every step in
our discussion of the Drude model, drastic approximations are made in order to keep the
theory as simple as possible — and yet the theory agrees very well with experiment and
explains many basic properties of metals and semiconductors.
        We conclude the chapter with some further, somewhat exotic, uses of solid state
theory. Quantum mechanics explains why solids are so difficult to compress, compared
to gases, and this effect, called degeneracy pressure, is explained in Section 13.10. In
Section 13.11, we will see that degeneracy pressure has profound consequences in
astrophysics. In the final two sections we describe a very strange state of matter that
occurs only at very low temperatures, the Bose-Einstein condensate.

13.2 Bonding of Solids
Solids, like molecules, are held together by several different types of bonds. Just as with
molecules, the distinctions between the various bonds are not always clear cut, and many
solids are bonded by a combination of several mechanisms. Nevertheless, we can
distinguish four main types of bonds — covalent, metallic, ionic, and dipole-dipole —
and we describe each of these in turn.


COVALENT BONDING
        We saw in Chapter 12 that many molecules are held together by covalent bonds,
each comprising two valence electrons shared by two of the atoms. In H2, for example, the
two hydrogen atoms share their two electrons to form a single bond; in CH4 (methane),
each of the four valence electrons of the carbon joins with an electron from one of the
four hydrogens to form a covalent bond. This same mechanism holds the atoms together
in many solids. A beautiful example of such a covalent solid is diamond, which is one of
the forms of solid carbon. In diamond each carbon atom is covalently bonded to four
neighboring carbon atoms. The four bonds from any one carbon atom are directed
symmetrically outward, and each atom in diamond is at the center of a regular tetrahedron
formed by its four nearest neighbors. [XX Need diagram of tetrahedral diamond
structure.] Both silicon and germanium — in the same group as carbon in the periodic
table — are bonded covalently in a similar tetrahedral structure. Other examples of
covalent solids are silicon carbide (widely used for its hardness and resistance to high
temperatures) and zinc sulfide (used by Rutherford and other pioneers in nuclear physics
as a scintillation detector).
        An important characteristic of a solid is its strength. One simple measure of this
strength is the atomic cohesive energy, defined as the energy needed, per atom, to
separate the solid completely into individual atoms. In Table 13.1 we have listed a
number of solids with their cohesive energies. The first three entries are covalent solids,
and we see what is generally true for covalent solids, that their cohesive energies vary
from around 3 to about 7 eV per atom. This energy scale is not surprising, since chemical
bonds involve the outer electrons in atoms, and a few eV is a typical energy for outer-



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       New Chapter 13, Solids I: Theory                                                 13.4

shell electrons. As one would expect, those solids with high cohesive energies, like
diamond and silicon carbide, are especially strong and have high melting points.
[TABLE 13.1XX]
        Since their electrons are usually well bound and cannot move easily, most
covalent solids are poor conductors of electricity. In many covalent solids, the excitation
energy of the electrons is higher than the energy of visible photons (2 or 3 eV). For this
reason many covalent solids (like diamond) cannot absorb visible photons and are
therefore transparent to light.


METALLIC BONDS
        As seen in Figure 13.1, more than half of all elements are metals, and all solid
metals are held together by the metallic bond. In some ways, this bond can be seen as a
relative of the familiar covalent bond. In a covalent bond, electrons are shared between
pairs of atoms. In the metallic bond, electrons are shared among all the atoms: One or two
valence electrons are completely detached from each parent atom and can wander freely
through the whole metal. Thus a metal can be thought of as a lattice of positive atomic
cores, immersed in a "sea" of mobile electrons, as in Figure 13.2. It is the attraction
between the positive atomic cores and this negative sea of electrons that holds the metal
together. As your can see in Table 13.1, the cohesive energy of typical metals is less than
that of many covalent solids, but is, nevertheless, high enough to make most metals rather
strong.
[FIG 13.1 XX]
[FIG 13.2 XX]
        All metals are good conductors of electricity because the outermost one or two
electrons per atom are free to move throughout the entire volume of the metal almost as if
they were free electrons. For this reason, they are sometimes described as a "gas" of
electrons. These free electrons are called conduction electrons, since they carry, or
conduct, electrical current. Since the conduction electrons can also carry energy, metals
are good thermal conductors as well. In addition, conduction electrons are the cause of
the opaque, but shiny appearance of metal surfaces. When a photon strikes a metal, it can
easily lose energy and momentum to the "free" electrons, and metals are therefore not
transparent to light. In the language of classical electromagnetism, when visible light
strikes the surface of a metal, the oscillating electric field of the light wave exerts an
oscillating force on the electrons in the metal. Since the conduction electrons are free to
move, they oscillate with large amplitude in response to the field, and their large
oscillations create a new electromagnetic wave, the reflected wave, which produces the
mirror-like appearance of metals.


IONIC BONDS
      We saw in Chapter 12 that certain pairs of elements can bond ionically to form
molecules. For exactly the same reasons, these same pairs of elements can bond ionically


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        New Chapter 13, Solids I: Theory                                                             13.5

to form solids. For example, in solid NaCl (common salt) each Na atom has lost one
electron to a Cl atom; the resulting closed-shell ions arrange themselves so that each Na+
ion is surrounded by several Cl– ions, and vice versa.* The resulting solid is therefore
quite tightly bound. As illustrated by the examples in Table 13.1, typical atomic cohesive
energies (energy to separate the solid into neutral atoms) are 3 or 4 eV per atom for ionic
solids.
        Since the electrons of an ionic solid are all bound tightly in closed-shell ions,
ionic solids are poor conductors of electricity and heat. For the same reason visible
photons cannot be absorbed easily, and ionic solids are transparent to light.


DIPOLE-DIPOLE BONDS
        The ionic bond results from the simple Coulomb force between oppositely
charged ions. If the basic units (atoms or molecules) are electrically neutral, this force is
not present. There is, nevertheless, a related, though weaker attraction that can act, even
between neutral atoms or molecules. A neutral atom or molecule can have its charges
distributed so that it has a electric dipole moment; that is, the preponderance of positive
charge is centered at one point and that of negative at another. Two neutral dipoles can
exert an electrostatic force on one another, as we now describe.
         Let us consider first two permanent dipoles, that is, two objects each of which is
a dipole even when in complete isolation. (An example of a permanent dipole is the water
molecule, as we saw in Chapter 12) If we bring these dipoles together so that two like
charges are closest, as in Figure 13.3 (a), the repulsive forces are slightly greater than the
attractive, and the net force is repulsive (See Problems 13.3XX and 13.4XX). On the
other hand, if the positive charge in one dipole is closest to the negative in the other, as in
Fig. 13.3(b), the net force is attractive. In addition to these forces, each dipole exerts a
torque on the other, and this torque tends to rotate them into the alignment of maximum
attraction (c). Therefore, the net force between two permanent dipoles (that are free to
rotate) is automatically attractive. Because it involves substantial cancellations between
attractive and repulsive forces, this dipole-dipole attraction is much weaker than the
Coulomb forces between the individual charges. Nevertheless, it is the attraction that
holds many solids together.
[FIG 13.3 = old 17.1 XX]
        Because of their basic spherical symmetry, no atoms are permanent dipoles. On
the other hand, many molecules (the so-called polar molecules, like H2O) are, and these
molecules can be bound into solids by the dipole-dipole attraction just described. Because
the bond holding the molecules to one another is much weaker than the bond holding the
atoms together inside each molecule, these solids are properly seen as composed of
molecules and are sometimes called molecular solids. An example of a molecular solid is
ice; we see from Table 13.1 that the energy needed to separate ice into H2O molecules is

        *
         Because each Na+ "belongs" to several Cl– ions, and vice versa, you should not think of solid
NaCl as made up of NaCl molecules; rather, it is a bonding of many Na atoms with many Cl atoms ,
forming one large, crystalline array.


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         New Chapter 13, Solids I: Theory                                                              13.6

about 0.5 eV per molecule. As expected, this is appreciably less than the cohesive
energies of typical covalent, metallic, or ionic solids (usually several eV per atom). It is
also much less than the 5 eV needed to remove any one atom from the H2O molecule —
which confirms that ice is properly regarded as an assembly of H2O molecules.
        The bond between two water molecules actually involves another effect as well.
We saw in Section 12.5 that the oxygen atom in water appropriates at least part of the
electron from each hydrogen, leaving nearly bare protons at the two ends of the molecule.
When water molecules bond to form ice, these protons are shared with nearby oxygen
atoms in somewhat the same way that electrons are shared in a covalent bond. This
bonding by shared protons is called hydrogen bonding and plays an important role in
biology: The DNA molecule comprises two intertwined helical strands of atoms. The
atoms within each strand are bound together by covalent bonds, while the binding
between the two strands is by hydrogen bonds, which are much weaker. This is what
allows the two strands to unravel from one another, without disturbing their internal
structure, in the process of replication.
        Although all atoms and many molecules have zero permanent dipole moment,
they can acquire an induced moment when put in the field of other charges. This is
because the field pushes all positive charges one way and pulls all negatives the other.
These induced dipoles are responsible for the very weak van der Waals bond* between
some atoms and molecules, as we describe shortly. Because this bond is so weak, it is
important only in solids where all other types of bond are absent. For example, the noble
gas atoms are subject to none of the bonding mechanisms described earlier, but can form
solids because of the van der Waals attraction. Because the bond is so feeble, the noble
gases solidify only at rather low temperatures. For instance, as shown in Table 13.1, solid
argon has a cohesive energy of only 0.08 eV/atom, and its melting point is 84 K. The van
der Waals force is also responsible for the solids of certain molecules that have no
permanent dipole moment, such as O2, N2, and CH4 (methane).
        The origin of the van der Waals force between two atoms (or molecules) that have
no permanent dipole moments can be explained by the following classical argument. A
classical atom consists of Z electrons orbiting around a fixed nucleus. The statement that
it has no permanent dipole moment means (classically) that, when averaged over time, the
dipole moment is zero. Nevertheless, as the electrons move around, they can produce an
instantaneous nonzero dipole moment. Now, consider two such atoms close together.
Suppose that at a certain instant atom 1 has a nonzero dipole moment. This produces an
electric field, which induces a dipole moment in atom 2. This induced dipole is always
aligned such that it is attracted by atom 1. (This is illustrated, for one possible orientation
of the atoms, in Figure 13.4. For other orientations, see Problem 13.5.) Therefore, as the
two dipoles fluctuate, each induces in the other a dipole that it attracts. This is the origin
of the van der Waals attraction between any two atoms or molecules.
[FIG 13.4 = OLD 17.2 XX]

         *
           The terminology here is a little confused. A few authors use "van der Waals" to describe both the
attraction between two permanent dipoles and the weaker attraction involving induced dipoles. We follow
the majority and reserve "van der Waals" to describe only the weaker induced dipole attraction.


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        New Chapter 13, Solids I: Theory                                                    13.7

13.3 Crystals and Noncrystals
All of the mechanisms that bond atoms (or molecules) together into solids have the same
general behavior: At large separations, r , the force between two bonding atoms is
attractive, at a certain separation, r  R , it is zero, and for r less than R it is repulsive.
This behavior is illustrated in Fig. 13.5, which shows the potential energy of two atoms as
a function of their separation r.
[FIG 13.5 = old 17.3 XX]
        When just two atoms interact as in Fig. 13.5, they have minimum possible energy,
and hence form a stable bound state, at separation R. Three such atoms would be in
equilibrium at the corners of an equilateral triangle, with the distance between each pair
equal to R. Similarly, four would be in equilibrium at the corners of a regular tetrahedron.
However, with five or more atoms, it is impossible to arrange them so that the distances
between all pairs are the same. Instead, the atoms must arrange themselves so that any
one atom is closest to a few others and farther away from all the rest. It is one of the tasks
of solid-state physics to predict, or at least explain, the arrangement of the many atoms in
a solid that minimizes their total energy and hence makes the solid stable.
        As one might expect, the stable configuration of many solids has the atoms
arranged in a regular pattern, in which the grouping of atoms is repeated over and over
again. This regular pattern is called a lattice, and the resulting solid is called a crystal.
Experimentally, the structure of crystals is found using diffraction of X rays, neutrons,
and electrons, and by the use of various kinds of microscopes (Figure 13.6). The
mathematical study of the possible crystal structures is an important part of theoretical
solid-state physics. Here we shall just describe a few such structures.
[FIG 13.6 = old 17.4 XX]
        The simplest crystal structure is the cubic lattice, in which the atoms are arranged
at the corners of many adjacent cubes, as shown in Fig. 13.7. In a perfect cubic lattice, all
these cubes are identical, and each is placed in the same relation to its neighbors.
Therefore, all the information about the structure of the lattice is contained in the single
cube, or unit cell, shown in Figure 13.7(c).
[FIG 13.7 = old 17.5XX]
         A slightly more complicated crystal lattice is the face-centered cubic (or FCC)
lattice, with atoms on each face of a cube as well as at the corners, as in Figure 13.8(a).
This arrangement actually gives a slightly higher density of atoms (more atoms per unit
volume) than the simple cubic, and is therefore favored in many solids. For example,
common salt (NaCl) is an FCC crystal. As shown in Figure 13.8(b), the Cl– ions are
arranged on one FCC lattice. The Na+ ions lie on a second FCC lattice, which is offset
from the Cl– lattice. In this way, each Na+ is nearest to six Cl– ions, and vice versa.
[FIG 13.8 = old 17.5 XX]




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         New Chapter 13, Solids I: Theory                                                                     13.8

EXAMPLE 13.1 Given that the nearest-neighbor separation in NaCl is r0  0.28 nm ,
estimate the atomic cohesive energy of NaCl in eV per atom.
        Let us consider first the potential energy of any one ion, for instance, the Na+ ion
at the center of the cube in Figure 13.8(b). This ion has six nearest-neighbor Cl– ions,
which give it a negative potential energy of 6ke 2 / r0 . However, the Na+ ion also has 12
next-nearest-neighbor Na+ ions at a distance of r0 2 . (These are all of the remaining 12
Na+ ions in the picture.) These contribute a positive potential energy of 12ke 2 / (r0 2) .
Next, there are eight Cl– ions (at the eight corners of the cube in the picture) at a distance
of r0 3 , and these contribute a negative potential energy of 8ke 2 / (r0 3) . Continuing in
this way we would find a total potential energy:
                                              ke 2     12   8              ke 2
               t ot al P E of any one ion        6          ...         .                         (13.1)
                                               r0       2    3              r0
The sum of the infinite series,  , is called the Madelung constant. Since the series is
difficult to evaluate, we shall simply state the result, that its sum is   1.75 (to three
significant figures). Therefore, the potential energy of any one ion in solid NaCl is
                                                          ke 2            1.44 eV  nm
                  t ot al P E of any one ion                 1. 75 
                                                           r0               0.28 nm                         (13.2)
                                                      9.0 eV.

In this calculation we have ignored the short-range repulsive forces due to the overlap of
the electron distributions when the ions get very close. The repulsive contribution to the
energy is actually rather small (for exactly the same reason as it was in ionic molecules;
see Section 13.2). Therefore, our estimate (13.2) is just a little below the observed value
                                 t ot al P E of any one ion  7. 9 eV.                                     (13.3)
For the remainder of this calculation, we shall use this observed value.
       To find the energy of the whole NaCl crystal, we cannot simply multiply the
answer (13.3) by the total number (N) of ions, since this would count each interaction
twice over.* Instead, we must multiply by N/2, to give a total potential energy of (-7.9
eV)N/2. Thus the energy to pull the solid NaCl apart into separated Na+ and Cl– ions is
(7.9 eV)N/2. Since N/2 is the number of Na+ - Cl– pairs, we can say that
     energy t o separat e NaCl int o Na + and Cl - ions  7.9 eV per Na + - Cl - pair
       If we wish to have separate neutral atoms, we must next remove an electron from
each C1– ion (which requires 3.6 eV), and then attach it to the Na+ ion (which gives us
back 5.1eV). Thus the total work done is


         *
            For each pair of ions, i and j, there is a single potential energy of interaction that went into (13.3).
However, if we simply multiply (13.3) by N, we will have counted the potential energy of i due to j and of j
due to i. In other words, we will have counted every interaction twice.


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         New Chapter 13, Solids I: Theory                                                                 13.9

   energy t o separat e NaCl int o Na and Cl at oms  ( 7. 9  3. 6  5. 1) eV per pair
                                                                     6.4 eV per Na-Cl pair.

Dividing by 2, we find finally that
                       atomic cohesive energy of NaCl  3. 2 eV/ atom.


[FIG 13.9 = old 17.7 XX]
        Many other solids have the FCC lattice, for example, diamond* and several
metals, including copper, silver, and gold. Two other important structures are the
body-centered cubic (BCC) and the hexagonal close packed (HCP), both sketched in
Figure 13.9. Many metals, such as sodium, potassium, and iron, are BCC; examples of
HCP are the solid forms of hydrogen and helium, and several metals, such as magnesium
and zinc.


MACROSCOPIC APPEARANCE OF CRYSTALS
         When a large enough number of atoms form into a single crystalline lattice, the
resulting solid is a macroscopic crystal. As the crystal forms, its macroscopic surfaces
tend to coincide with the planes of its microscopic lattice; if a grown crystal is struck, it
tends to cleave along these same atomic planes. Either way, the crystal acquires smooth
flat surfaces that reflect light like a mirror. If the material is transparent, these same
surfaces can cause strong internal reflections, which add to the glittering appearance that
we associate with crystals. There are many examples of such macroscopic crystals:
ordinary table salt, gem stones, the fluorite crystal in Figure 13.10, and many more.
[FIG 13.10 = old 17.8 XX]
        Many solids, including most metals, have a microscopic crystalline structure but
do not, nevertheless, appear crystalline. This is because, instead of forming as a single
large crystal, these solids form in a jumbled array of many tiny microcrystals, called
grains, as illustrated in the micrograph of stainless steel in Figure 13.11. Since a typical
microcrystal, or grain, is about 1  m across, the crystalline nature of such solids is
completely invisible to the naked eye. Nevertheless, each grain contains some 1010 atoms,
and, from a microscopic point of view, the dominant feature of the material is its
crystalline structure.
[FIG 13.11 = old 17.9 XX]



         *
           This claim does not contradict our earlier statement that diamond has a tetrahedral structure. The
carbon atoms are placed on two FCC lattices, one offset from the other. Each atom on either lattice is
covalently bonded to its four nearest-neighbor atoms, all of which are on the other lattice at the corners of a
regular tetrahedron.




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       New Chapter 13, Solids I: Theory                                                    13.10

EXAMPLE 13.2 An important task for solid-state physics is to predict the strengths of
materials. Given the estimate in Problem 13.22 for the force needed to separate two
bonded atoms, make a rough estimate of the tensile strength of a simple cubic material
with lattice spacing r0  0.3 nm .
The tensile strength of a material is the force per unit area needed to rupture it. From
Problem 13.22XX we find that the force needed to separate a single pair of bonded atoms
is of order F  3  109 newt on . The density of atoms in any one crystal plane is 1/ r0 .
                                                                                          2


Therefore (if we ignore all but nearest-neighbor interactions), the force to tear apart a unit
                                     2
area of two adjacent planes is F / r0 ; that is

                                         F              3  109 N
                 t ensile st rengt h                       10
                                                                      3  1010 N/ m 2 .   (13.4)
                                                      (3  10 m)
                                              2                    2
                                         r0
This calculation assumed a cubic lattice and included only nearest-neighbor interactions.
Furthermore, although the force that we used is realistic for strong bonds (covalent, ionic,
or metallic), it would be a gross overestimate for weaker bonds such as the van der Waals
bond. Nevertheless, the answer (13.4) should — and does — give the right order of
magnitude for any single crystal of strongly bonded atoms. Single-crystal whiskers of iron
(as in Figure 13.12) and of quartz have tensile strengths of this order. Macroscopic
samples of many crystalline materials have tensile strengths much smaller than (13.4)
(that for cast iron is some 100 times smaller), but this is because such macroscopic
samples are not perfect crystals. It is the many imperfections in these materials that break
easily and explain their much lower tensile strengths.
[FIG 13.12 = old 17.10 XX]


AMORPHOUS SOLIDS
        Not all solids have a crystalline structure. Instead, the atoms of many solids, such
as glass, wax, and rubber, arrange themselves in an irregular pattern. Such solids are
described as amorphous, from a Greek word meaning "without form." In fact, many
materials can solidify in either a crystalline or an amorphous form, depending on their
preparation. For example, a material may form in a crystalline state if cooled slowly from
a liquid, a process called annealing. (Slow cooling gives the atoms time to arrange
themselves in the lowest-energy, crystalline, state.) On the other hand, the same material
may form in an amorphous state if quenched, that is, cooled very rapidly.
         The difference between crystalline and amorphous solids is shown schematically
in Figure 13.13. Part (a) illustrates a perfect crystal, with its atoms in a rigid and regular
lattice. Part (b) shows an amorphous solid, with its atoms bonded in a more-or-less rigid
structure, but not in a perfectly regular pattern. For comparison, parts (c) and (d) illustrate
a liquid and a gas. When one heats a solid sufficiently, it first melts [part (c)]: Although
its atoms remain within the range of interatomic forces, they are no longer held in a rigid
pattern and are, instead, free to move around. When heated still further, the liquid




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         New Chapter 13, Solids I: Theory                                                            13.11

becomes a gas [part (d)]: The atoms separate much further, they move more rapidly, and
they interact only during their brief collisions.
[Fig.13.13 = old 17.11 XX]
        We can characterize the differences between crystals, amorphous solids, liquids,
and gases in terms of their long-range order. A perfect crystal has perfect long-range
order: If we know its crystal structure and the location of one atom, we can predict with
certainty the location of all other atoms. This is illustrated in Figure 13.14(a), which
shows the probability density (the probability of finding an atom) plotted along one
crystal axis for an ideal crystal. If we know that there is an atom at the origin, we can
predict with certainty that there are atoms at r0 , 2 r0 , 3 r0 , . . . , for as far as the solid
extends.*
[Fig.13.14 = old 17.12 XX]
        The distribution function for an amorphous solid is like Figure 13.14(b). If we
know that there is an atom at the origin, the probability of finding another atom very close
to r  0 is small, since two atoms repel one another strongly at short range. The
probability increases to a peak at about the same r0 at which the next atom of a crystal
would be located. It then drops again because the repulsion between overlapping atoms
prevents another atom from locating too close to the atom of the first peak. Because of the
amorphous solid's irregular structure, we can predict the atoms' positions with less and
less certainty as we move on, and the density function rapidly approaches a constant,
reflecting our complete ignorance of the positions of the more distant atoms. We say that
the amorphous solid shows short-range order, but no long-range order.
        The probability density for a liquid is almost exactly like that for an amorphous
solid, shown in Figure 13.14(b). In fact, an amorphous solid can be regarded as liquid,
"frozen in time." The probability density for a gas is shown in part (c). It is much lower
than those for solids or liquids, since the density is much less; it shows even less structure
than that for the amorphous solid because a gas has even less order.

13.4 Energy Levels of Electrons in a Solid; Bands
We have stated that the reason metals conduct electricity is that some of their conduction
electrons can move essentially freely within the boundaries of the metal, but this does not
explain why the electrons in a metal can move so freely, nor why the electrons in an
insulator cannot. To explain this difference we must examine the energy levels for the
electrons in a solid. Fortunately, we can do this quite easily, leaning on our discussion of
the energy levels of electrons in molecules (Section 12.4). As one might expect, once we



         *
           Strictly speaking, Fig. 13.14(a) represents a classical crystal at absolute zero. In quantum
mechanics the uncertainty principle means that we cannot know the locations with complete certainty, and
at nonzero temperatures thermal motions introduce a further uncertainty. For both reasons the ideal spikes
of Fig. 13.14 (a) should be a little spread out.




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         New Chapter 13, Solids I: Theory                                                              13.12

understand the energy levels of the electrons, we can explain many properties of solids in
addition to their conductivity,
        In Section 12.4 we described the energy levels of an electron belonging to either
of two atoms as the atoms come together to form a molecule. While the atoms are far
apart, we can focus on any one atomic orbital (1s, 2s, . . . ). Since the electron can occupy
this orbital in either atom, this gives two degenerate orbitals (that is, two possible wave
functions with the same energy). We saw in Section 12.4 that as the two atoms come
closer together and begin to overlap, this degeneracy splits, and we have instead two
orbitals of different energy, as indicated in Figure 13.15(a).
[FIG 13.15 = old 17.13 XX]
        If we start instead with three atoms far apart, then for each atomic orbital there are
three independent, degenerate wave functions for an electron (corresponding to attaching
the electron to any of the three atoms). As we move the atoms together this threefold
degeneracy is split, and we get three distinct possible energies, as indicated in Figure
13.15(b). More generally, if we consider N atoms far apart, then for each atomic orbital
there are N degenerate wave functions, and as we move the atoms together these fan out
into N more-or-less evenly spaced levels. We have shown this in Figure 13.15(c) where
we have indicated the many closely spaced levels by shading.
        We can imagine putting together a whole solid in this way, starting with N
well-separated atoms. While they are still far apart, we can arrange the atoms in the same
geometrical pattern as the final solid, and we can then imagine the whole structure slowly
contracting to its equilibrium size. As the nearest-neighbor separation r decreases, the
energy levels of the electrons will vary, their general behavior being that shown in Figure
13.15(c), which we have redrawn in more detail as Figure 13.16. For each atomic orbital
there are N independent wave functions, all of which have the same energy when r is
large. As we reduce r, this degenerate level fans out into N distinct levels. We refer to
these N closely spaced levels, which all evolved from a single atomic level, as a band.
The width and spacing of these bands varies with r, but the bands of the actual solid are
found by setting r equal to its equilibrium value r0 , as shown on the left of Figure 13.16.
[FIG 13.16 = old 17.14 XX]
        Figure 13.16 introduces some of the terminology used in connection with the
energy levels of electrons in solids. The band of levels that evolved from the 1s atomic
levels is called the 1s band, and similarly with 2s, 2p, and so on. The range of energies in
a single band — the width of the band — is typically a few eV. If there are N levels
within the band,* the average spacing  E between adjacent levels is
                                                    a few eV
                                            E              .
                                                        N



         *
          As we shall see shortly, there are generally more than N states in a band, but this only strengthens
our conclusion that the average spacing is very small.


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         New Chapter 13, Solids I: Theory                                                               13.13

In practice, N is very large; even in a microcrystal 1  m across, there are some 1010
atoms (Problem 13.25XX). Thus the level spacing within a band is very small; in a
microcrystal with N  1010 ,  E is of order 10–10 eV, and in a larger crystal it is even
smaller. Therefore, for almost all practical purposes, we can think of the electron energy
levels in a solid as continuously distributed within each band. This is what the shading of
the bands in Figure 13.16 was intended to suggest.
        It can happen that the highest level in one band is above the lowest level in the
next-higher band. In this case we say that the two bands overlap. On the other hand, it
often happens (as in Fig. 13.16) that two neighboring bands do not overlap, and there is
instead a band gap — a range of energies that are not allowed. We shall see that the
occurrence of band gaps, and their widths, are two of the main factors that determine the
conductive properties of a solid.
         It is important to know whether all the states in a given band of a solid are
occupied (just as it was important in our discussion of the periodic table to know whether
all the states of a given level in an atom were occupied). To this end, we need to know
how many states there are in a given band. For s bands (the 1s and 2s bands of Figure
13.16, for example) there are N independent wave functions and two possible spin
orientations, and hence 2N states in all. If we consider a p band (a band that evolved from
an l  1 atomic level), then when the N atoms were well separated there were 3N
independent wave functions (three different values of lz for each atom) and two spin
orientations. Therefore, there are altogether 6N states in a p band. Since we shall only be
concerned with s and p bands, we shall not give a general formula here (but see Problem
13.28XX).

13.5 Conductors and insulators – a qualitative view
We are now ready to explain the difference between conductors and insulators. We
illustrate our discussion with three examples, starting with lithium, which we know to be
a metal and hence a conductor.


LITHIUM: A CONDUCTOR
        The lithium atom has the configuration 1s22s1, with its 1s level full and its 2s
valence level half full. As we assemble N lithium atoms into solid Li, the 1s and 2s levels
fan out into the 1s and 2s bands, which we have shown in Figure 13.17. The N lithium
atoms contribute altogether 3N electrons, while each of the two bands can hold 2N
electrons. Thus the 1s band is completely filled, and the 2s valence band is half filled. We
have shown this occupancy in Figure 13.17, where dark shading represents occupied, and
light shading unoccupied levels. The energy shown as E F is the Fermi energy, which is
defined as the energy of the highest occupied level.*

         *
           Strictly speaking, this is correct only at absolute zero. At higher temperatures, thermal motions
can excite a few electrons to levels somewhat above EF as describe in Section 13.10.




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       New Chapter 13, Solids I: Theory                                                  13.14

[FIG 13.17 = old 17.15 XX]
         We are now going to argue that the electrons in the partially full valence band can
move around in response to an applied electric field — they are conduction electrons, so
the partially full valence band is also called the conduction band — whereas the
electrons in the full 1s band cannot move. To see this, let us consider a lithium wire
oriented along the x axis. In the absence of an electric field, the energy levels are the same
at all points along the wire, as indicated in Figure 13.18(a).
[FIG 13.18 = old 17.16 XX]
         Suppose now that we connect a battery to the ends of the wire to maintain an
electric field to the right. This field raises the potential energy of the electrons on the
right, and the bands now look like Figure 13.18(b) or, in a magnified version, Figure
13.18(c). For any electron at the top of the occupied levels in the conduction band, there
are now unoccupied levels of the same energy (and lower) just to the left. Therefore, each
of these electrons begins to move to the left in response to the electric field, as indicated
(for one electron) by the horizontal arrow in (c). This motion of electrons to the left is, of
course, the expected electric current. In due course, the electrons moving to the left
collide with defects and other "imperfections" of the lattice and lose some kinetic energy,
as indicated by the downward arrow in Figure 13.18(c). These collisions, which are
discussed in Section 13.8, are the main cause of electrical resistance in most metals. The
sequence of accelerations and collisions is continually repeated as the electrons move
along the wire. The energy lost by the electrons in collisions increases the vibrations of
the lattice, and the wire's temperature rises. This temperature rise is just the familiar Joule
heating of a wire by a current.
        The argument just given does not apply to electrons in the full 1s band, since there
are no vacant states to their immediate left. These electrons are immobilized by the
complete lack of empty states to transfer into. Thus the electrons in a filled band do not
contribute to conduction.
        This example already illustrates our two most important conclusions: Solids with
a partially filled band are conductors; those in which all occupied bands are 100% full are
insulators. Our next two examples show that these simple rules have to be applied with
care.


BERYLLIUM: A CONDUCTOR
        We turn now from lithium to the next element in the periodic table, beryllium. Its
atomic configuration is 1s22s2, and both of its occupied levels are filled. This might
suggest that solid Be should have filled 1s and 2s bands and should therefore be an
insulator, whereas Be is, in fact, a good conductor. The explanation of this apparent
contradiction is not hard to find: As we assemble solid Be and its atomic levels fan out
into bands, the 2s and 2p bands spread so far that they overlap one another, as shown in
Figure 13.19. The 2s and 2p bands thus form a single valence band, which can
accommodate altogether 8N electrons. (Remember an s band holds 2N electrons and a p



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       New Chapter 13, Solids I: Theory                                               13.15

band 6N.) Thus, of the 4N electrons in the solid, 2N fill the 1s band, and the remaining 2N
can only partially fill the valence band. Therefore, Be is a conductor.
[FIG 13.19 = old 17.17 XX]


DIAMOND: AN INSULATOR
         Atomic carbon has the configuration 1s22s22p2, with its n  2 , valence shell half
filled. This suggests that solid carbon should be a conductor, which, indeed, it is in the
form of graphite. However, solid carbon in the form of diamond is an outstanding
insulator. To explain this puzzle, we must look carefully at the behavior of the energy
levels as we assemble a diamond. In diamond the carbon atoms are covalently bonded. As
we bring the atoms together and the covalent bonds form, the 2s and 2p levels fan out into
bands, as we would expect. However, each level splits into two separate bands, as shown
in Figure 13.20, (This is closely analogous to the development of the bonding and
antibonding orbitals described in Section 12.4.) At the equilibrium separation r0 , the
lower two bands overlap each other, as do the upper two. Thus the band structure of
diamond is as shown on the left of Figure 13.20: There is an n  1 band that can hold 2N
electrons, and then two well-separated n  2 bands, each of which can hold 4N electrons.
Since carbon has 6N electrons in all, the lowest two bands are completely full, and the
upper n  2 band is completely empty. Because the lower n  2 band holds all the
valence electrons, it is called the valence band, and because electrons in the upper n  2
band (if there were any) would contribute to conduction, this band is called the
conduction band. With its valence band full and its conduction band empty, diamond is
an insulator. It turns out that as indicated in Figure 13.20, the gap between the valence
and conduction bands of diamond is about 7 eV. Under normal conditions, this means
that no electrons can be excited into the empty conduction band. (At room temperature,
for example, thermal motions could excite an electron, but the probability is so low that
the expected number of excited electrons in a large diamond is far less than 1.) This
makes diamond an extremely good insulator. This same large band gap also explains why
diamond is transparent. Visible photons have energies of 2 to 3 eV and cannot excite any
electrons across the 7 eV gap; therefore, visible light cannot be absorbed in diamond.
[FIG 13.20 = old 17.18 XX]
        We mentioned in Section 13.2 that solid silicon (just below carbon in the periodic
table) has the same crystal structure as diamond. The 3s and 3p levels in Si behave almost
exactly like the 2s and 2p levels of diamond, and silicon, like diamond, is a
nonconductor. There is, however, a subtle but crucial difference. The equilibrium
separation r0 in Si is larger than in diamond (0.24 nm in Si compared to 0.15 nm in
diamond). It is easy to see in Figure 13.20 that a larger value of r0 suggests a smaller
band gap, and, indeed, the band gap in silicon is only about 1 eV (compared with 7 eV in
diamond). Since visible photons can easily excite electrons across a 1 eV gap, light is
quickly absorbed in silicon, which is therefore not transparent. The narrow band gap in
silicon also has a dramatic effect on its conductivity. At low temperatures, all the valence
electrons in Si are locked in the full valence band, and Si is an insulator. But as the


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         New Chapter 13, Solids I: Theory                                                            13.16

temperature rises, a substantial number of electrons are excited across the narrow gap into
the conduction band, and the conductivity increases rapidly. For this reason, silicon and
other solids with a similar narrow gap between a full valence band and an empty
conduction band are called semiconductors. Semiconductors include silicon and
germanium, both from group IV of the periodic table, and several compounds of group III
with group V, such as gallium arsinide. Semiconductors and their enormous technological
importance are discussed in greater detail in Chapter 14.

13.6 The Drude Model of Conductivity
In this and the following two sections, we look more closely at the conductivity of metals.
We describe a simple but powerful model of electrical conduction, called the Drude*
model. The Drude model is an example of a "semi-classical" model, meaning it is a
hybrid of classical and quantum ideas. In the Drude model, we think of conduction
electrons as classical particles with a definite position and velocity, but to understand the
interaction of these electrons with their surroundings, we use the results of quantum
calculations. The classical view of electrons as point particles can be made roughly
consistent with the quantum view of electrons as de-localized wavefunctions through the
notion of wavepackets, described in section 6.7XX. A wavepacket describing an electron
in a metal can have both an average position and an average velocity, and so, within the
constraints set by the Heisenburg uncertainty principle, we are allowed to view electrons
in a metal as localized particles. See Problem 13.XX.
        Before describing the Drude model, we begin with some definitions. The
electrical conductivity  of material is defined by the equation
                                                 j  E .                                            (13.5)
Here, E is the electric field† and j is the current density, which is defined as the current I
per area A, that is, j = I / A. Both the current density and the electric field are vectors,
but, since we will be working in one dimension, we will drop the vector notation and use
signs to represent direction, with (+) meaning rightward and (–) meaning leftward. In the
SI system, conductivity has the units of (ohmm)-1, although one often sees the units of
(ohmcm)-1 used. The resistivity  of a material is defined as the inverse of the
conductivity,   1/  , so resistivity has units of ohmm or ohmcm. Equation (13.5)
can then be rewritten as
                                                 E j .                                             (13.6)

This equation is the microscopic version of Ohm's Law V = IR , which is discussed in
Sec13.8. In Problem 13.X, you will show that (13.6) is indeed equivalent to Ohm's Law.




         *
             Pronouced ―DREW-dah.‖
         †
         Here we use E for electric field, whereas, in earlier chapters, we used  to avoid confusion with
the symbol for energy.


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       New Chapter 13, Solids I: Theory                                                  13.17

        A good electrical conductor, like copper, has a large  and a small  , while a
good electrical insulator, like glass, has an extremely small  and a huge  . Table 13.1
and Figure 13.21 shows the conductivities and resistivities of various materials. Notice
that conductivity varies over some 24 orders of magnitude — an enormous range — from
107 or 108(ohmm)–1 for good metals to 10–16 (ohmm)–1 for insulators.
[FIG 13.21 New XX]
        In 1900, just 3 years after J. J. Thompson discovered the electron and well before
the development of quantum mechanics, Paul Drude (German, 1863-1906) proposed a
model of electrical conduction in metals, a model which is still used today because of its
simplicity and usefulness. Drude made the bold assumption that the electrons in a metal
behave as a gas of free particles. These free electrons, which we now understand are the
conduction electrons, accelerate when an electric field is applied. If there were no
frictional forces present, then the electrons would accelerate without limit, leading to an
ever-increasing current. Since such runaway currents are never observed, there must be
some mechanism at work which acts to slow the electrons. Drude thought that the
electrons must collide with the nuclei of atoms, but this turns out to be incorrect. Because
of a quantum mechanical effect, which we shall describe later, the electrons do not collide
with the nuclei but instead they collide with both "impurity atoms" and "lattice
vibrations." These terms will be explained in detail in the next section.
        Regardless of what the electrons collide with, we can define a scattering time  as
an average time between collisions. Between collisions, an electron with charge q = e
in an electric field E will experience a force F = q E = – e E and hence an acceleration a =
F/m = – e E/m, where m is the mass of the electron. If the electron starts from rest, then,
in a time t, the electron will acquire a velocity
                                                   eE
                                      v  at           t.                             (13.7)
                                                    m
Here, the minus sign indicates that, for negatively charged electrons, the direction of the
velocity is opposite to the direction of the electric field. The effect of collisions, whatever
their origin, is to momentarily interrupt this free acceleration and decrease, stop, or
reverse the forward velocity. Our picture of the electron motion is then one of a very
jerky, start-stop kind of motion. In the competition between the speeding up due to the E-
field and the slowing down due to collisions, the electron acquires an average velocity
called the drift velocity, vd . With equation (13.7) in mind, we now define the scattering
time  by the relation
                                                 eE
                                        vd        .                                  (13.8)
                                                 m
The scattering time is the average time between collisions. A given electron actually
experiences a wide range of time intervals between successive collisions, and properly
defining an "average" time interval has some surprising subtleties which are explored in
Problem 13.xx.




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       New Chapter 13, Solids I: Theory                                                 13.18

       Drude related this scattering time  to the electrical conductivity  by the
following argument. First, we show that the current density j can be written as
                                          j  n q vd .                                  (13.9)

where n is the concentration (number per volume) of charge carriers, q is the charge per
carrier, and vd is the drift velocity. To understand this equation, consider a wire carrying a
current I, as in Figure13.22. In a time t , all the charges will move a distance
x  vd t as shown. A cylindrical portion of the wire of length x and cross-sectional
area A contains a total charge
[FIG 13.22 new XX]
             Q = (charge per carrier)  (number of carriers)
                 = (charge per carrier)  (number per volume)  (volume)
                 = q n ( x A )  q n (vd t A ).

All the charges making up this total charge Q will pass through the right end of the
cylinder during a time interval t , and the resulting current is I  Q / t . The current
density is then
                                       I   Q
                                 j             q n vd ,
                                       A   t A

and we have demonstrated (13.9).
       Now inserting (13.8) into (13.9), and setting q = – e, we obtain
                                       eE     ne 2 
                              j  nq 
                                                  E .                            (13.10)
                                      m       m 
Comparing (13.10) with the equation j =  E, we conclude that


                                                ne 2
                                                                                     (13.11)
                                                 m


This equation is the central result of Drude's model of conductivity. It says that the
conductivity is large when there are many charge carriers per volume (large n), few
collisions (large  ), and light charge carriers (small m). The dependence on mass can be
understood intuitively by noting that low-mass carriers acclerate more quickly in an
applied field ( a  qE / m ), leading to a larger  . Consequently, ionic conductors such as
electrolytic solutions, in which massive ions carry the current, have a much smaller
conductivity than normal metals, in which light electrons are the charge carriers, as seen
in Table 13.2.



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       New Chapter 13, Solids I: Theory                                                  13.19

Example13.3. A copper wire of diameter 1.0 mm (radius r = 0.5 mm) carries a current
of I = 10 A. Estimate the magnitude of the drift velocity vd.
According to (13.9), j  n q v d , the drift velocity is easily computed if the current density
j and the carrier concentration n are known. The current density is readily calculated as
                  I   I          10 A
            j                              1.27  105 A/ m 2  12. 7 A/ cm 2 .
                  A r  2
                             (5  103 m) 2
(Here, we give the number in both SI units, as well in the more practical units of A/cm2,
often used by electrical engineers.) Computing n, the number density of conduction
electrons in copper, is not quite so easy. One can compute it roughly by noting that, for
most solids, the nearest-neighbor atoms are about r0 = 0.3 nm apart and for copper, the
valence or number of conduction electrons per atom is 1, as one might surmise by noting
its electronic configuration — a single s electron outside a d shell. We conclude that the
number density of charge carriers for copper is about
                                    1             1
                             n                            4  1028 m -3 ,
                                   r03
                                         ( 0.3  109 m) 3
(The measured value is actually a factor of 2 higher than this.) Finally, we can compute
the drift velocity as
                  j              1.27  105 A/ m 2
         vd                          3          19
                                                        2  105 m/ s  0.02 mm/ s .
                nq        ( 4  10 m )(1.6  10 C)
                                  28



This is a very tiny speed; a given electron in this wire takes more than a day to travel a
meter! When you flip a light switch, the electromagnetic signal travels along the wire at
nearly the speed of light, but the electrons themselves move with glacial slowness. A long
garden hose filled with water behaves similarly. When you open the faucet, water comes
out the far end of the hose immediately, due to the incompressibility of water, while a
given water molecule takes several seconds to travel the length of the hose.

13.7 Electron Collisions in Metals
We now return to the nature of the collisions experienced by conduction electrons in a
metal. Understanding the origin of these collisions requires a knowledge of quantum
mechanics, which did not exist in Drude's time.
        In a metal, the atomic nuclei surrounded by their inner core electrons form an
array of positively charged ion cores. A conduction electron sees these ion cores as an
array of attractive potential wells separated by barriers, as shown in Figure 13.23. To
understand how an electron interacts with these barriers, we must recall some qualitative
features of quantum mechanical tunneling, which was discussed in Chapter 7.
[FIG 13.23 new XX]
       When a free electron wave packet encounters a single potential barrier, as in
Figure13.24(a), there is, in general, a reflected wave and a transmitted wave. If the barrier



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       New Chapter 13, Solids I: Theory                                                 13.20

height U is high enough or the barrier width L is sufficiently large, the electron will be
almost entirely reflected. If there are two barriers, as in Figure13.23(b), then a
remarkable phenomena called resonant tunneling can occur. If the energy of the incident
electron wave has certain special values, then the waves reflected from the two barriers
will interfere destructively, resulting in no net reflected wave and 100% transmission of
the incident wave. There is an analogous process in the field of optics: an anti-reflection
coating on a glass surface. If the coating has the correct thickness and refractive index,
then an incoming wave of a particular frequency experiences 100% transmission and no
reflection.
[FIG 13.24 new XX]
        If instead of two barriers, there is an infinite array of uniformly spaced identical
barriers, as in Figure 13.23, then it turns out that resonant transmission occurs over a
range of electron energies. In this band of energies, the electron behaves essentially as a
free electron. It can travel forever, without scattering. This range of energies in which
free propagation occurs is precisely what we called the conduction band in Section 13.5.
         If the nuclei in a metal formed a perfectly regular array, then, because of resonant
tunneling, the conduction electrons could zip right through the metal unimpeded, and the
metal would have infinite conductivity. However, any disruption in the periodicity of the
lattice causes electron scattering, and there are two causes of non-periodicity which are
always present in a metal. First, any metal sample always contains some impurities, no
matter how carefully it has been purified. Impurities may be foreign atoms, or they may
be defects, such as vacancies (missing atoms), interstitial atoms, or grain boundaries, all
of which are illustrated in Figure 13.25.
[FIG 13.25 new XX]
         Besides impurities and other defects, lattice vibrations also disrupt the periodicity
of the lattice and cause electron scattering. At any nonzero temperature, the atoms of a
solid vibrate about their equilibrium positions, as in the ball-and-spring model of a solid
shown in Figure 13.26. The higher the temperature, the more vigorous the vibrations. At
a high enough temperature, this "jiggling" of the atoms becomes so great that the solid
melts. In the correct quantum mechanical description of this ball-and-spring model of a
solid, this vibrational energy is quantized and the quanta of energy are called "phonons".
In classical language, higher temperature means larger amplitude vibrations, while in the
language of quantum mechanics, higher temperature means more phonons, and
consequently more electron-phonon scattering. It turns out that at room temperature and
above (T 300K), the rate of electron-phonon scattering is proportional to temperature,
leading to a resistivity proportional to temperature.
[FIG 13.26 new XX]
        In Figure13.27, we indicate schematically the potential that a conduction electron
encounters in three environments: (a) a perfectly periodic potential such as is almost
acheived in very pure metals at very low temperatures; (b) the non-periodic potential
seen in a metal with some impurities, in which case different types of atoms produce
different well-depths; and (c) the non-periodic potential found in a pure sample at high


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        New Chapter 13, Solids I: Theory                                                  13.21

temperatures, in which case lattice vibrations produce a non-uniform spacing of the
atoms. In cases (b) and (c), compared to case (a), there is much more scattering of
electrons, a shorter time  between collisions, and a smaller conductivity  = ne2 /m.
[FIG 13.27 new XX]
[FIG 13.28 new XX]
        The general shape of the curves of resistivity vs. temperature found in metal
samples are displayed Figure13.28. Three curves are shown corresponding to three
different samples:
(a) A "clean" sample, such as moderately pure copper or aluminum. In this case, at
higher temperatures (room temperature and above), lattice vibrations are the dominant
source of electron scattering leading to a linear dependence of  on T. At very low
temperatures (a few ten's of Kelvin or lower) lattice vibrations are so weak that impurities
dominate electron scattering, leading to a temperature-independent resistivity.
(b) An "ordinary" sample, very like (a) but not so pure. Here, the impurity scattering is
larger than in case (a) leading to a higher resistivity at low T. The dependence of the low-
temperature resistivity on purity leads to a simple way to quantify the purity of samples:
the residual resistivity ratio (RRR) or "triple-R" is defined as the ratio of the resistivity at
room temperature to that a helium temperatures (a few Kelvin). The higher the triple-R,
the more pure the sample. Carefully purified metal samples have a triple-R of a few
hundred; very dirty samples have a triple-R of less than 2.
(c) A "dirty" (very impure) metal or an alloy. An alloy is a mixture of metals; an example
is brass, which is a mixture of copper and tin. In an alloy, almost every atom can be
regarded as an "impurity" and consequently the impurity scattering is so large that it
dominates the resistivity at all but the highest temperatures. Another example in this
category is stainless steel, which is iron with a very high concentration of other elements
including chromium and nickel. Stainless steel can be regarded as an alloy or as very
"dirty" iron .
        In Figure 14.29, we plot the resistivity  against temperature T for a typical metal
such as copper and for a pure semiconductor, such as silicon. Drude's equation
  ne 2 / m gives us insight into the very different behaviors of metals and
semiconductors seen in this graph. In metals, the temperature dependence of the
conductivity is due entirely to the temperature dependence of the scattering time . The
carrier concentration n is temperature-independent, while the scattering time  decreases
with increasing temperature. As the temperature increases,  decreases due to increased
scattering from lattice vibrations; consequently, the conductivity decreases and the
resistivity rises,. In semiconductors, however, the temperature dependence of the
resistivity is due mostly to the temperature dependence of the carrier concentration. In
semiconductors, n is small compared to metals, and very strongly temperature-dependent.
The smallness of n results in a large resistivity compared to metals. In pure (undoped)
semiconductors, the carrier concentration n is thermally activated: as the temperature is
increased, more and more electrons are thermally excited from the valence band into the



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       New Chapter 13, Solids I: Theory                                                 13.22

conduction band, across the band gap; n increases rapidly with increasing temperature,
leading to a decrease in resistivity.
[FIG 13.29 new XX]

13.8 The Fermi Speed
A very important feature of the conductivity of metals, a feature we have not yet
discussed, is that the conductivity  in the equation j   E is independent of the applied
electric field E or the current density j. That is to say, although j and E may vary
considerably, the ratio j/E remains constant. Conductors which exhibit a field-
independent conductivity are called ohmic conductors. Resistors made from ohmic
conductors obey Ohm's Law, which is the statement that V/I = R = constant, where I is
the current through the sample, V is voltage across it, and R is its resistance. It is
important to note that the equation R = V/I , by itself, is not Ohm's Law; it is merely the
definition of the resistance R of a sample. Ohm's Law is the statement that the resistance
(defined as V/I) is a constant, independent of V or I. Most conductors are ohmic, so long
as the current density j is low enough that Joule heating does not raise the temperature of
the sample. If significant Joule heating does occur (as in most incandescent light bulbs),
then the increased temperature causes an increase in  due to increased electron-phonon
scattering, and the sample under study exhibits non-ohmic behavior.
        The fact that the conductivity of a metal is independent of the applied electric
field E is surprising for the following reason. The conductivity depends on the scattering
time  according to the Drude formula (13.11),   ne 2 / m . At first glance, it might
seem that the scattering time should depend on the field E. After all, if the field E is
increased, the acceleration a of a conduction electron will increase (according to a =
qE/m), and if the electron accelerates more quickly it will surely encounter an impurity or
phonon more quickly, resulting in a shorter collision time  and a smaller conductivity  .
This argument, which implies that  depends on E, would be correct, if conduction
electrons had zero initial velocity after a collision. However, due to a quantum
mechanical effect which we explain shortly, the electrons in a metal always have a very
large intrinsic speed, called the Fermi speed vF. The Fermi speed is about 0.3% of the
speed of light ( vF  0.003c  106 m/ s ). In contrast, the magnitude of the drift velocity vd
, which is the mean velocity due to the applied field E, is miniscule ( vd  0.01 mm/ s , as
in Example 13.1). Consequently, the increase in velocity due to the E field (and hence
the decrease in time until the next collision) is completely negligible compared to the
initial huge Fermi speed of the conduction electron, and the scattering time is almost
completely independent of the field E. This lack of dependence of  on E is the origin of
Ohm's Law.
        The motion of conduction electrons is thus one of extremely rapid, random,
motion due to the large Fermi speed and, superimposed on this, a very small bias toward
one direction due to the tiny drift velocity. As shown in Figure 13.30, a conduction
electron in a metal follows a random zig-zag path even with zero applied field (E = 0),
regardless of the temperature. This is an example of random walk diffusive motion ,


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        New Chapter 13, Solids I: Theory                                                           13.23

which was described in Sec.3.8. The mean free path* l is defined as the mean distance an
electron travels between collisions, that is, the step length in the random walk. The mean
free path is related to the Fermi speed and the scattering time  by
                                                       l
                                                vF        .                                     (13.12)
                                                       
The mean free path (mfp) depends on the purity and temperature of the sample. Very
dirty samples, such as alloys, or metals at very high temperatures, may have a mfp of
nanometers or less. Metals of ordinary purity at room temperature have a mfp of roughly
10-100 nm. While high-purity (―clean‖) samples at low temperatures may have a mfp of
many micrometers or even millimeters.
[FIG 13.30 new XX]


Example 13.4. Using the Drude model and the known Fermi speed, estimate the
scattering time and the mean free path for conduction electrons in copper at room
temperature.
Drude's equation,   ne 2 / m , allows us to compute the scattering time, , from the
measured conductivity, , given in Table 13.2. From our previous Example (13.3), the
conduction electron density is about n = 8  1028 m–3. Thus,
                m          (9.11  1031 kg)[5.9  107 (   m) 1 ]
                                        3          19
                                                                       3  1014 s .
                n e2           (8  10 m )(1.6  10 C)
                                      28                       2



Assuming that the Fermi speed is approximately 106 m/s, we have for the mfp
         l  vF   (106 m/ s)(3  1014 s)  3  108 m  30 nm .

This is approximately 100 lattice constants. (A lattice constant is defined as the linear
size of a ―unit cell‖ in a crystal; it is roughly equal to the distance between nearest-
neighbor atoms.)
        We now consider why conduction electrons in metals have such a large intrinsic
speed, the Fermi speed. We first note that the Fermi speed is not simply the thermal speed
of the electrons. As described in Section 3.7, in a classical gas of particles of mass m at
temperature T, the average thermal speed v of the particles is given by the
expression 1 mv 2  2 kT . For electrons at room temperature, this thermal speed is about
            2
                     3


105 m/s, a factor of 10 smaller than the Fermi speed. Another difference between the
thermal and Fermi speeds is that the Fermi speed is independent of temperature, while the
thermal speed rises with increasing temperature.
       The large Fermi speed of conduction electrons is a consequence of the Pauli
exclusion principle. To see why, we begin by noting that conduction electrons in a metal

        *
          In Section 3.8, we used the symbol  for the mean free path. Here, we use the symbol l, to avoid
confusion with the de Broglie wavelength.


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       New Chapter 13, Solids I: Theory                                                  13.24

can be regarded as free electrons in a box, a problem which was considered in Chapters 7
and 8. Of course, the electrons are not truly free, that is, they are not in a region where
the potential energy function U(x) is constant. An individual electron experiences the
attraction of the ion cores of the atoms in the sample, as well as a repulsive potential
energy due to the other conduction electrons. However, the Coulomb repulsion from the
other electrons nearly cancels the Coulomb attraction to the ion cores, and the electrons
behave as a "nearly free" particles.
        Let us imagine placing electrons in an empty box, one at a time. Fig.13.31 shows
the wave functions of the various electron states for the case of a one-dimensional box of
length L. The first two electrons will occupy the ground state (n = 1) with one electron
spin up and the other spin down. Because of the Pauli exclusion principle, the next two
electrons will be forced to occupy the first excited state (n = 2); the next two will occupy
the 2nd excited state (n = 3), etc. As more electrons are poured into the box, they are
forced to occupy higher and higher energy states. This filling of states is the partial filling
of conduction band states described in Section 13.5. If there are very many electrons in
the box (many electrons in the conduction band) then the average energy of an electron
will be quite large compared to thermal energies. This is the reason that conduction
electrons in a metal are so energetic. In a metal, the density of conduction electrons is
such that all of the quantum states up to a high energy level are occupied. As we said in
Section 13.5, this highest occupied level is the Fermi level, and its energy is called the
                         2
Fermi energy EF = 1 mv F . (As you might guess, all this was first explained by the Italian-
                    2
American physicist Enrico Fermi.) In a metal, the Fermi energy has a value of a few
electron volts, and, this corresponds to a speed, the Fermi speed, of about 106 m/s.
[FIG 13.31 new XX]
       We can estimate the value of the Fermi energy by the following argument.
Notice, in Figure13.31, that the nth state of the system has a wavelength  which satisfies
                                              
                                          n       L.                                  (13.13)
                                              2
When there are N electrons in the system, the highest occupied state is the state with n =
N/2, so we can write
                                              
                                          N       L.                                  (13.14)
                                              4
If we temporarily regard electrons as classical (rather than quantum) objects, then the
average distance between nearest neighbor electrons is r = L/N, which by eqn.(13.14),
becomes r =  / 4 . Although we have made this argument in one dimension (1D), a
similar result is true in 3D; that is, in either 1D or 3D, the average distance between
electrons is very roughly equal to the wavelength of the highest occupied state. In 3D, a
careful treatment gives r   / 2 , rather than the 1D result, r   / 4 . We can now
compute the kinetic energy of electrons at the Fermi energy. We write




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          New Chapter 13, Solids I: Theory                                                  13.25

                        p2     h2       h2                              h2
                    E                                                                 (13.15)
                             2 m 2
                                                          
                                                                2
                        2m            2m 2r                           8 m r2

where we have taken r   / 2 as the nearest neighbor distance. If we consider a
monovalent metal, that is, a metal with one conduction electron per atom, then the
average distance r between conduction electrons is the same as the average distance
between atoms, which is about 0.3 nm for most solids. We then have for the Fermi
energy,
                           h2                ( 6.6  1034 J  s)2
                  EF           
                                  8   9.1  1031 kg    0.3  109 m 
                                                                            2
                         8 m r2                                                           (13.16)
                                    19
                          7  10         J  4 eV


The Fermi energy varies from metal to metal due to variations in the valence of the metal
and the density of the atoms, but it always falls in the range from 2 to 12 eV. This is the
same order of magnitude as the energy of chemical bonds, so the result is not too
surprising. We can now, finally, compute the Fermi speed. Since E F  1 m vF2,
                                                                            2



     vF          2E F / m    2   7  1019 J  /  9.1  1031 kg        1.2  106 m/ s.   (13.17).

Note, once again, how enormous this speed is compared to the drift speed.
        We conclude our discussion of the Drude model with the story of a curious
historical accident. As we have mentioned before, Drude's work was done well before
the advent of quantum mechanics, so there was no way for Drude to compute the Fermi
speed or the mean free path. Why then was Drude's model immediately successful? It
was because Drude's calculations contained two large mistakes which cancelled! From
the known values of the conductivity of metals, Drude used his formula (13.11) to
correctly compute the scattering time  . As a check on the theory, he also computed the
scattering time using  = l/v, where l is the mean free path and v is the mean electron
speed. But his values for l and v where both too small by an order of magnitude. He
assumed, incorrectly, that the electrons scattered from atomic nuclei, so his estimate of
the mean free path l was about a nanometer, smaller by a factor of at least 10 than the
correct value for metals at room temperature. He also incorrectly assumed that the speed
of the electrons is given by the thermal speed of about 105 m/s, which is about 10 times
smaller than the correct Fermi speed. Because both numbers were wrong by about the
same factor, the ratio,  , came out correct. Consequently, the model appeared to be
nicely self-consistent and it was immediately accepted by physicists. More than 30 years
passed before enough quantum mechanics was known for the correct picture to emerge.
Today, we remember Drude because his boldly wrong model leads to the correct
formula,   ne 2 / m . In science, unlike politics, the good that physicists do lives after
them, while the bad is oft interred with their bones.



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              New Chapter 13, Solids I: Theory                                                         13.26

13.9 Degeneracy Pressure

    This material will be used in the next section, Section 13.10, but will not be needed after that.
In the previous sections we have described the properties of conduction electrons in
metals. Because these electrons behave as if they were nearly free, they are often referred
to as a "gas" of electrons. However, the behavior of this gas is quite unlike that of an
ordinary gas of molecules, such as air. The electron gas is an example of a degenerate
fermi gas — called degenerate because the system is near its ground-state*, and fermi
because it consists of fermions, that is, particles which obey the Pauli exclusion principle
as explained in Section 10.5. One of the properties of a degenerate fermi gas is that it is
extremely hard to compress. If you squeeze on a block of metal, like copper, you find that
to compress it significantly requires roughly 1 million atmospheres of pressure. This
incompressibility of metals is largely explained by a quantum mechanical effect called
degeneracy pressure, which occurs in degenerate fermi gases.
        The basic physics of degeneracy pressure can be understood by considering the
simplest possible fermi gas, namely a single electron, confined to a rigid 1D box, and in
its ground state. The ground-state wave function is shown in Figure 13.32. If the box has
length L, the wavelength of the ground state is  = 2L and the ground state energy is,
from (13.15) E  h 2 / (8mL2 ) . If the box is compressed, decreasing L, then the kinetic
energy of the ground state must increase since E proportional to 1/L2 . (Recall that in
quantum mechanics, shorter wavelength corresponds to higher kinetic energy.) So
compressing the container and decreasing the wavelength, necessarily increases the
energy. Since the energy is increased, work must be done to perform the compression;
the gas is hard to compress. The electron wants to lower its energy by spreading out over
a larger volume; that is, it wants to push the walls of the enclosure outward. This
pressure which the confined electron exerts on the walls of its container is called the
degeneracy pressure.


[FIG 13.32 new XX]
        This argument is readily extended to many electrons (or fermions) in a 3D box.
The n quantum state of an electron in a 1D rigid box has wavelength  such that
         th

n  / 2  L and its energy is

                                         p2    h2       h2
                               En                         n2                                      (13.18)
                                         2m   2m 2   8 m L2

In a 3D cubical box of volume V = L3, the energy becomes

              *
            There are two distinct uses of the term "degenerate" in quantum mechanics. The first use was
described in Section 8.3: Two wavefunctions are degenerate if they have the same energy. Now we
encounter a different use: A many-particle system is degenerate if the system is near its ground state so that
its particles are "crowded" into a limited number of single-particle states; the system is non-degenerate if
the temperature is high enough that the system is far from its ground state and its particles have many empty
single-particle states to choose from.


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       New Chapter 13, Solids I: Theory                                                                       13.27

                                    h2
                                       2  x
                               E         n 2  n y2  n z2                                                 (13.19)
                                   8m L

where nx, ny and nz are the quantum numbers labeling the state. For many non-interacting
electrons in a 3D box, the total energy is the sum of the one electron energies
                                           h2
                            E t ot 
                                         8 m L2
                                                    n
                                                    i
                                                            2
                                                            xi    n yi  n zi 
                                                                     2      2
                                                                                                             (13.20)

where the sum is over all occupied states i. At low temperatures, each level up to the
Fermi level will be occupied with 2 electrons. Now a marvelous simplification occurs. In
our present analysis, only the dependence of the energy on the dimensions of the
container matters. So we rewrite the total energy (13.20) as
                                                   A      A
                                   E t ot           2
                                                               ,                                            (13.21)
                                                   L     V 2/ 3
where A is a constant and V = L3 is the volume of the cubical box.
        We can use this simple expression for the total energy to compute the pressure of
the Fermi gas. Recall from thermodynamics, that for a gas of volume V and pressure p, if
the volume of a gas is changed by an amount dV, and no heat is added or removed (a so-
called adiabatic change), then the work done on the gas is equal to the change in energy
                                 dW  dE   p dV .                                                          (13.22)

The minus sign here indicates that the energy increases when the volume decreases. The
pressure exerted by the gas on the walls of its container is thus
                                  dE
                         p                   (adiabat ic condit ions) .                                    (13.23)
                                  dV
Combining (13.21) and (13.23), we find that the degeneracy pressure is
                       d  A                      5 / 3                  2 / 3       1       2 E grd
             p                   
                                          2
                                              AV                 2
                                                                      AV            V                   ,   (13.24)
                      d V V 2 / 3 
                                          3                       3
                                                                                                 3 V
where, in the last step, we have used (13.21) and have assumed that the temperature is
sufficiently low that the total energy Etot is approximately the ground state energy Egrd.


Example 13.4. Estimate the degeneracy pressure of an electron gas with a concentration
equal to that in an ordinary metal.
We use equation (13.24) and our earlier estimate (13.16) of the Fermi energy, EF  4 eV.
For a degenerate Fermi gas, the particles occupy single-particle states with energies that
range from zero to EF. We therefore expect that the average energy per particle is roughly
EF/2 (the exact answer turns out to be (3/5)EF) . The total energy of a gas is the sum of the
energies of all the particles in the gas. So, if there are N particles in the gas, the total
energy is E grd  N E F /2, and equation (13.24) becomes



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        New Chapter 13, Solids I: Theory                                                      13.28

                                  2 E grd   1 N EF             1
                           p                                  n EF                     (13.25)
                                  3 V       3 V                3
where n = N/V is the concentration of electrons (number/volume). For a metal, this is
roughly n = 1/a3, where a  0.3 nm is a typical lattice constant, and so the pressure is
               EF         ( 4 eV)  (1.6  10-19 J / eV)
        p                                                 1011 N/ m 2  106 at m.
               3a 3           (3)  (0.3  10-9 m)3
        (13.26)
(In the last step, we have used 1 atmosphere = 1 atm  105 N/m2.) This is an enormous
pressure, comparable to the pressure at the center of the Earth. The reason metals do not
blow apart is because this pressure is balanced by the very strong Coulomb attraction
between the negative electrons and the positive ion cores.
        Degeneracy pressure makes degenerate fermi gases much harder to compress than
ordinary molecular gases. If you squeeze on a molecular gas, such as air, at STP
(standard temperature and pressure: T = 273 K, p = 1 atm) with a pressure of two
atmospheres, its volume will decrease by about a factor of two This is easy to deduce
from the ideal gas law; see Problem 13.xx. In contrast, to substantially compress a solid
such as a block of copper requires a pressure of the same order of magnitude as the
degeneracy pressure — about 106 atm. (See Problem 13.xx.). Degeneracy pressure has
some remarkable consequences for the evolution of stars, as we shall see in the next
section.

13.10 White Dwarfs, Neutron Stars, and Black Holes
                                                                                th
 This section, though it describes some of the most remarkable discoveries of 20 century
astronomy, contains material that will not be used again and can be omitted without loss of
continuity.
The most spectacular examples of degenerate fermi gases occur in the realm of
astrophysics. Degeneracy pressure profoundly influences the evolution of stars and leads
to fantastistically dense stellar remnants called white dwarfs and neutron stars. An
ordinary star, like the Sun, has a long, stable "middle age" during which the star shines
because the temperature and pressure near its core are high enough for hydrogen nuclei
(protons) to fuse to form helium. This fusion reaction produces the energy which makes
the star shine, and it also produces an outward pressure which conteracts the inward pull
of gravity, so the star has a stable size. However, eventually the supply of hydrogen near
the star's core is depleted to the extent that the fusion reaction slows and the core begins
to collapse because the inward pull of gravity is unopposed. Details of the subsequent
evolution depend on the mass of the star, but story is usually violent. As the core
collapses, it becomes hot enough that helium begins to fuse to form even heavier
elements, and the outer layers of the star are blown off by the violent reactions in the core.
Eventually, the nuclear fuel is exhausted and the star further collapses under its own
gravity. If the star is not too much more massive than the sun, it will collapse until
electron degeneracy pressure (13.24) grows large enough to counteract gravity. The star
then is extremely dense, with a mass comparable to the Sun's mass packed into a sphere


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        New Chapter 13, Solids I: Theory                                                  13.29

the size of the Earth, as in Figure 13.33. Such a star is called a white dwarf, because,
initially, it is hot enough that it is classified as "white hot". But, over time, the star cools
by radiation and becomes a "black dwarf", a cold, immensely massive, cinder. Countless
white dwarf stars have been identified by astronomers; they are common objects in the
sky.
[FIG 13.33 new XX]
         In 1937, a young theoretical physicist, S. Chandresakhar, discovered that if a star's
mass is greater than 1.4 solar masses — the so-called Chandresakar limit — even
electron degeneracy pressure is insufficient to stop to the gravitational collapse. As the
star collapses into an object smaller than a white dwarf, a process called inverse beta
decay occurs; at sufficiently high pressures, an electron will burrow into a proton,
forming a neutron and a neutrino: p  e  n  ν . (See Section 17.4, for discussion of
closely related reactions.) The star is then composed primarily of neutrons. Neutrons, like
electrons and protons, are fermions; they obey the Pauli exclusion principle and can exert
degeneracy pressure just as electrons do. If the star's mass is less than about 5 solar
masses, then neutron degeneracy pressure will balance gravity when the diameter of the
star is about 10 km. This fantastically dense object is called a neutron star; the density
at its center is about 1018 kg/m3, which is the density of "nuclear matter" in the interior of
atomic nuclei. The escape speed at the surface of a neutron star is about 0.8 c, 80% of the
speed of light. Problem 13.XX explore the calculation of the diameter of a neutron star.
        Another remarkable feature of neutron stars is their rapid rate of rotation. As a
star collapses, it must spin faster, due to conservation of angular momentum — just as a
skaters spins faster as they pulls in their arms. Sun-sized stars typically have rotational
periods of days to months; after collapse into an object of 10 km diameter, a neutron star
spins at rates of 10 to 1000 times per second. The rapid rotation of the neutron star and
its magnetic field interacts with intersteller dust and gas nearby to produce a bright light
source which rotates with the star, causing the star to appear to flash on and off as seen
from the earth. This rapidly flashing neutron star is called a pulsar, and, since the 1960's,
several dozen pulsars have been positively identified.
[CHANDRESAKAR BIO]
        If the mass of the collapsing star is greater than roughly 5 solar masses, then even
neutron degeneracy pressure is insufficient to counteract the inward pull of gravity and
the star collapses into an infinitely dense, infinitesimally small object called a singularity.
At present, we have no understanding of physics in the vicinity of a singularity; all our
theories (relativistic quantum mechanics, general relativity) break down under such
conditions. Surrounding the singularity is an imaginary sphere called the event horizon
which marks the boundary at which the escape speed equals the speed of light. The event
horizon around a 10-solar-mass singularity is a few kilometers in diamter. No light and no
signal of any kind can escape from inside the event horizon; things can fall in, but nothing
can come out. The singularity and its surrounding event horizon are therefore called a
black hole. The evidence for the existence of black holes is indirect but very strong. We
can "see" black holes by the effect of their gravity on their surroundings. It is believed



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         New Chapter 13, Solids I: Theory                                                             13.30

that super-massive black holes, of some 106 solar masses, reside at the center of most
galaxies, including our own Milky Way. See Figure 13.34.
[FIG 13.34 new XX]



13.11 Classical and Quantum Gases

 This material will be needed in the following section and in Chapter 15, Statistical Mechanics, but not
elsewhere in this text.
The most general definition of a "gas of particles" that one can give — a definition
sufficiently broad that it is valid in the framework of either classical or quantum
mechanics — is this: A gas is a collection of non-interacting or weakly-interacting
particles. In this section, we look closely at the distinction between classical and
quantum gases. A classical gas is one that is well described with classical mechanics,
while a quantum gas is one in which quantum effects are important. We have already
seen in Section 13.9 one example of a quantum gas, namely, the degenerate fermi gas
formed by conduction electrons in a metal. We will see that a gas behaves classically
when conditions are such that the particles of the gas are far apart, compared to their
wavepacket diameters. However, for a gas of identical particles, if the particles overlap,
then quantum effects cannot be ignored. The properties of a quantum gas depend
crucially on whether the particles of the gas are fermions or bosons. Before describing
quantum gases, let us review the properties of fermions and bosons, which were briefly
discussed in Section 10.5.
        All particles can be classified as either fermions or bosons. A fermion is defined
as a particle which obeys the Pauli exclusion principle. Examples of fermions are the
electron, the proton, and the neutron. In Section 10.4, the Pauli exclusion principle was
stated as: "No two electrons in a quantum system can occupy the same quantum state".
We can state this principle more generally as: "No two identical fermions can occupy the
same quantum state".
        Bosons are defined as particles which do not obey the Pauli Exclusion. In a
quantum system consisting of many identical bosons, an unlimited number of the bosons
can exist in the same quantum state. One example of a boson is the alpha particle, which
is the nucleus of a helium atom consisting of two protons and two neutrons. Another
example is the deuteron, which is a nucleus with one proton and one neutron. Note that
the term "particle" is not limited to "elementary" particles such as electrons, but includes
complex bound collections of particles. For instance, a neutral atom of rubidium-87,
consisting of 37 protons, 37 electrons, and 50 neutrons, can be regarded as a "particle"
which happens to be a boson. The deuterium atom, which is an isotope of hydrogen
consisting of an electron (a fermion) bound to a deuteron (a boson), is a fermion.
        How does one tell if a particle is a boson or a fermion? One way, in principle, is
to test whether the particle obeys the Pauli Exclusion Principle; that is, put many of the
particles in a box and see if more than one can occupy a given quantum state. If only one
particle can occupy each state, then the particle is a fermion; if more than one can, then it


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        New Chapter 13, Solids I: Theory                                                   13.31

must be a boson. Fortunately, there is an easier way to distinguish between fermions and
bosons — the spin of the particle determines its type. We saw in Chapter 9 that electrons
have an intrinsic angular momentum called the spin. All particles, no matter how
complex, have a spin, that is, an internal total angular momentum. Like energy, the
internal angular momentum of a particle or of a system of particles is always quantized;
the maximum z component of the angular momentum can always be written as s , where
the number s, which is called the spin, is either an integer (0, 1, 2, 3,...) or a half integer
(1/2, 3/2, 5/2, etc.). Experimentally, it is found that if the total spin is an integer, then the
particle is a boson; if the spin is half-integer, then the particle is a fermion. (This
correspondence between spin and the Pauli principle can be derived from relativistic
quantum theory.) Electrons, protons, and neutrons all have spin 1/2, so they are fermions.
Composite particles, such as atoms, can be either fermions or bosons. In general, if an
atom contains an even number of fermions, then it is a boson. For instance, in the ground
state of the rubidium-87 atom, the spins of the 37 protons, 37 electrons, and 50 neutrons
add up in such a way that the angular momenta of all (37+37+50 = 124) spin-half
fermions cancel and produce a boson with spin zero. If an atom has an odd number of
fermions, then its total spin is half-integer and the atom as a whole is a fermion.
         If a collection of identical particles is placed in a rigid box or some other quantum
potential well, then which quantum states are occupied by the particles depends both on
the temperature and on whether the particles are fermions or bosons. At lower
temperatures, the particles tend to occupy the available states with the lowest energy,
while at high temperatures, the particles tend to spread out in energy and occupy states of
both low and high energy with nearly equal probability. In Figure 13.35, we compare the
behavior of a gas of fermions and a gas of bosons. The figure shows the spectrum of
energy levels of some quantum system, such as a rigid box or a harmonic oscillator or
some other, more complex, potential. These levels are called "single-particle states"
because they are the solutions of the time-independent Schrödinger equation for any one
particle. If we assume that the potential energy function [U(x, y, z)] in the Schrödinger
equation does not involve the spin of the particle, then the levels are degenerate; for the
case of an electron, each level is at least doubly degenerate and represents two quantum
states, one with spin up and one down.
[FIG 13.35 new XX]
        We now imagine filling these quantum states with identical particles, either
bosons or fermions. We assume that the particles are non-interacting, so that the presence
of other particles does not affect the potential which a given particle experiences, and the
solutions of the one-particle Schrödinger equation remain valid, even though there is
more than one particle present. At absolute zero (T = 0 K), the particles will occupy the
lowest available states. For the case of spin 1/2 fermions, the Pauli exclusion principle
requires that each orbital be occupied by no more than two fermions, and so the states are
filled up to a maximum level, the Fermi level. However, for bosons, the Pauli Principle
does not apply; all the particles can happily occupy the same quantum state, and at T=0,
the bosons all crowd into the lowest energy state, the ground state. As we have seen in
previous sections, examples of degenerate fermi gases include conduction electrons in
metals and in white dwarf stars, as well as neutrons in neutron stars. Degenerate fermi


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       New Chapter 13, Solids I: Theory                                                13.32

gases are common. In contrast, the degenerate boson gas, also called a ―bose gas‖ or
"bose-einstein condensate", is a very exotic state and is found only at extremely low
temperatures. As far as we know, degenerate bose gases do not occur in nature. A
human-made example of a degenerate bose gases is described in the next section.
        Still looking at Figure 13.35, we now imagine raising the temperature of our gas
slightly. As explained in Chapter 3, at temperature T, a particle has a thermal energy of
about kT, where k is Boltzmann's constant. This results in a thermal smearing of the
occupancy of the levels. For fermions, some of the states above the Fermi level are now
occupied, due to thermal activation of the particles, leaving some of the states below the
Fermi level vacant. For bosons, not all the particles are in the ground state; some occupy
excited states about one kT above the ground state.
        Now let us raise the temperature considerably. With the temperature high enough,
then, regardless of whether the particles are fermions or bosons, they are now widely
dispersed among the quantum states. The average occupancy of every state is much less
than one, and it is unlikely that more than one particle will compete for occupancy of the
same state. In this high temperature regime, a collection of particles behaves as a
classical gas, because quantum effects due to multiple occupancy are negligible; in this
classical regime, a gas of bosons behaves just like a gas of fermions. In contrast, when
the temperature is sufficiently low that the occupancy of the states depends on whether
the particles are fermions or bosons, then we are in the degenerate regime. In this low-
temperature regime, the collection of particles is referred to as a quantum gas.
        The temperature required to reach the classical regime depends on the density of
the particles in the system. The gas is in the classical regime if the density is low enough
so that particles can "stay out of each other's way", that is, if their quantum mechnical
wave packets do not overlap. We can estimate the temperature and density necessary for
this classical behavior by the following argument, which has 3 main ingredients:
(1) If the number of particles per volume is n, and the average distance between
neighboring particles is r, then, on average, the volume per particle is r3, and
                                                1
                                          n       .                                  (13.27)
                                                r3
(2) A free particle with kinetic energy E and momentum p has a deBroglie wavelength 
related to E and p by
                                          p2   h2
                                    E            .                                  (13.28)
                                          2m 2m 2


(3) For a free particle at temperature T, the kinetic energy is given by
                                    E  1 m v 2  3 kT .
                                        2         2                                   (13.29)

where v is the mean thermal speed. This relation was discussed in Section 3.7.
[FIG 13.36 new XX]


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       New Chapter 13, Solids I: Theory                                                           13.33

       Wave packets have a size of at least the DeBroglie wavelength , so the condition
of non-overlapping wavepackets can be written a   ; that is, the mean distance
between particles is much greater than their wavelength, as seen in Figure 13.36. We can
combine points (2) and (3) [equations (13.28) and (13.29)] to obtain an expression for the
deBroglie wavelength:
                                         h2                             h
                            3
                                kT                    or                     .                (13.30)
                                        2m 2
                            2
                                                                       3m k T

 This last expression is called the thermal DeBroglie wavelength; it is the wavelength of
particles at a temperature T. At high temperatures, the DeBroglie wavelength is small,
and the particle can be represented by a small, compact wave packet which behaves very
much like a classical particle. At very low temperatures, however, the wavelength is
large, and the wave packets describing the particles may overlap or even fill the volume
of their container.
        Finally, we combine observation (1) [Eqn. (13.27)] with (13.30) to write the
classical-gas condition a   as
                                                                                      3/ 2
                      1                 h                                 3m k T 
                                              , or          n                2             .   (13.31)
                  n 1/ 3               3m k T                             h      
We have achieved our goal. This last equation is gives the conditions of number density
and temperature required for a gas to be in the classical regime.
       We can define a quantum concentration nq as
                                                                3/ 2
                                               3m k T 
                                         nq      2                                            (13.32)
                                               h      
which is the approximate particle concentration at the crossover from classical to
quantum regimes. At concentrations higher than this, the wavepackets overlap strongly,
quantum effects cannot be ignored, and the gas is in the degenerate quantum regime. At
concentrations much less than this, the gas is in the classical regime and the ideal gas law
applies.


Example 13.5 What are the thermal deBroglie wavelengths of free electrons and
neutrons at room temperature?
At room temperature, T  300 K. From equation (13.30) we have, for electrons

         h                                   6.63  1034
                                                                              6.2  109 m  6 nm
        3m k T             3   9.11  10   31
                                                     1.38  10   300 
                                                                23


.
For neutrons,


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           New Chapter 13, Solids I: Theory                                                          13.34

              h                                   6.63  1034
                                                                                  1.5  1011 m  0.015 nm
             3m k T           3  1.67  10     27
                                                          1.38  10   300 
                                                                      23


.
Note that for electrons, the room-temperature wavelength is much greater than the
distance between atoms in a solid. Thus, for conduction electrons in a solid, the wave
packets strongly overlap, and they must form a degenerate quantum gas.


Example 13.6. Consider the degenerate fermi gas formed by conduction electrons in a
metal. How high must the temperature be raised before the gas enters the classical
regime?
The crossover from degenerate fermi gas to non-degenerate classical gas occurs when the
number density n is equal to the quantum concentration nq given by (13.32). Solving the
equation n = nq for the temperature T, we obtain
                                                         h 2 n 2/ 3
                                            T                                             (13.33)
                                                          3m k

Since there is about 1 conduction electron per atom in a metal, the number density of
conduction electrons is roughly the same as the density of atoms, and we have n = 1/a3 ,
where a is the lattice constant, the distance between nearest-neighbors atoms in the metal.
For all solids, the lattice constant is roughly a = 0.3 nm . We now substitute n = 1/a3 into
(13.33) and obtain
                     h2
           T 
                   3m k a 2
                                                                                           (13.34)
            
                                      6.6  10 
                                                34 2

                                                                              130000 K
                3   9.1  1031     1.4  10    0.3  10 
                                                  23                 9 2



This temperature is far above the melting point of any solid. We conclude that for any
solid metal, the temperature is always well below that needed to excite its electrons into
the classical regime. Conduction electrons are always in the strongly degerate quantum
regime.



13.12 Bose-Einstein Condensation

    This material can be omitted without loss of continuity.
In 1924, Einstein predicted that a gas of identical bosons will exhibit very strange
behavior if cooled to a low enough temperature. Building on earlier work by the Indian
physicist S. Bose (after whom bosons are named), Einstein showed that below a certain
critical temperature, most of the bosons in a gas will crowd into the ground state. This



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         New Chapter 13, Solids I: Theory                                                              13.35

low-temperature collection of ground-state bosons is called the Bose-Einstein
condensate.
         Before the work of Bose and Einstein, it was already well understood that the
particles in a gas of bosons will all enter the ground state when the temperature is very
close to absolute zero. Particles in a gas have a thermal energy of about kT, and so we
expect that most of the particles will be in the ground state when the temperature is so
low that this thermal energy is less than the energy of the first excited state in the system.
This temperature turns out to be very, very, very low -- about 10–13 K for an ordinary gas
(see Problem 13.XX). What Einstein showed was that bosons will condense into the
ground state at a significantly higher temperature than this. The Bose-Einstein
condensation occurs when the particles’ wavefunctions begin to overlap, namely, at the
temperature of crossover from the classical to degenerate quantum regimes described in
the previous section. This critical temperature, given approximately by equation (13.33),
depends on the density of the gas and the mass of the particle, but for a low-pressure gas,
it is typically around 10–8 K or higher (See Problem 13.XX). Einstein’s derivation is a bit
too subtle to describe here, but what he showed was that the phenomenon of bose
condensation is a consequence of the symmetry of the wavefunction of indistinguishable
particles (described in Section 10.5). He showed that not only can bosons occupy the
same state, but that below a certain critical temperature, they actually prefer to.
        More than 70 years passed before the 1924 prediction of Einstein was verified
experimentally. In 1995, two physicists at the University of Colorado, Carl Wieman and
Eric Cornell, led a team of researchers who acheived Bose-Einstein condensation (BEC)
in a gas of rubidium-87 atoms. In 2001, Cornell and Wieman, along with MIT physicist
Wolfgang Ketterle, received the Nobel Prize for their experimental studies of the Bose-
Einstein condensation*.
[Cornell and Wieman BIO]
       We briefly describe the original experiment of Wieman and Cornell, which is a
marvel of ingenuity. The basic problem was how to cool a gas sample down to 10-7 K, a
temperature that had not been achieved before. They began by suspending the gas sample
in a magnetic trap so that the atoms of the gas did not touch the hot walls of the sample
container. They then cooled the gas in two stages. In the first stage, a technique called
laser Doppler cooling was used to cool a sample of about 1017 atoms down to 10-5 K. In
laser Doppler cooling, lasers shine on the sample from six directions (Figure 13.37). The
frequency of the laser light is tuned to be near an electronic transition of the atom, so that
when the atom is moving toward the laser light, the Doppler-shifted frequency of the light


         *
            Yet another example of experimentalists winning a Nobel Prize for verifying a theoretical
prediction due to Einstein. In 1926, French experimentalist Jean Perrin won the Prize for verifying
Einstein’s theory of Brownian motion. In 1923, American experimentalist Robert Millikan won the Prize
for verifying Einstein’s theory of the photoelectric effect. In 1993, American radio astronomers Russell
Hulse and Joseph Taylor won the Prize for their studies of gravitational radiation from a radio pulsar, which
verified Einstein’s theory of General Relativity. Einstein’s theory of Special Relativity is at the foundation
of the field of High-Energy Physics, which has produced more than two dozen Prizes. Einstein won the
Prize once, in 1921, for the photoelectric effect.


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       New Chapter 13, Solids I: Theory                                                 13.36

is just right to cause an inelastic collision that slows the atom. Because of the Doppler
shift, the atom only interacts with the laser beam it is heading into. In this way, the atom
is slowed by one of the six beams, no matter which way it travels, and it gradually cools
by transferring its energy and momentum to re-emitted light.
[FIG 13.37 new XX]
        In the second stage of cooling, the temperature is lowered by controlled
evaporation. The magnetic container is adjusted so that the fastest-moving (hottest) atoms
have just enough energy to escape the sample, leaving behind cooler-than-average atoms.
When the original sample of 1017 atoms was allowed to evaporate down to 2000 atoms,
the temperature had dropped to 10–7 K and the Bose-Einstein transition occurred. (See
Figure 13.37 and Problem 13.XX). Virtually all the rubidium atoms dropped into the
ground state, and for all practical purposes, the sample was at absolute zero. This
experiment was especially elegant because all the equipment was inexpensive and small
enough to fit on a table-top. Cheap diode lasers and hand-wound magnetic coils were
used, rather than the million-dollar, whale-sized, liquid helium cryostats and
superconducting magnets found in most low-temperature laboratories.
[FIG 13.38 new XX]
         The work of Wieman, Cornell, and Kitterle provided the first experimental
verification of Bose-Einstein condensation in a gas, that is, a collection of weakly-
interacting particles. However, the basic phenomena had already been observed in
systems of strongly-interacting particles, the two clearest examples being
superconductivity and superfluidity. Superconductivity, which is described in more
detail in the next chapter, is a phenomenon which occurs in many metals. When cooled
to a few degrees above absolute zero, superconducting metals loose all electrical
resistance and become perfect electrical conductors. The conduction electrons in these
metals interact with the lattice vibrations in a rather complex way and combine to form
bound pairs of electrons that have total spin zero and are thus bosons. It is these electron-
pairs that undergo a Bose-Einstein-like transition at a few degrees Kelvin and cause
superconductivity. Superfluidity is a phenomenon which occurs in liquid helium. The
most common isotope of helium, 4He, is a boson. Helium forms a liquid at temperatures
below 4.2 K; below 2.2 K, liquid helium-4 undergoes a Bose-Einstein transition to form a
so-called superfluid, which can flow through a pipe with zero friction. (See Problem
13.XX) The amazing properties of superconductors and superfluids are a consequence of
a large fraction of the particles being in the quantum ground state.
        Although superconductors and superfluids are both fairly well understood, a
detailed theoretical description has proved enormously difficult and elusive, because of
complications arising from the strong interactions between particles. The original theory
of Bose-Einstein condensation provides a very clear theoretical picture, but really only
applies to the simple case of a gas of non-interacting particles. Now that BEC has finally
been achieved in a gas, a clean comparison between theory and experiment is possible,
and rapid progress is being made in this re-vitalized field.




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         New Chapter 13, Solids I: Theory                                                           13.37



CHECK LIST FOR CHAPTER 13
         CONCEPT                                     DETAILS
Types of bonding in solids                  Covalent, metallic, ionic, dipole-dipole
Crystals and non-crystals                   Long-range order in crystals, short-range order only in amorphous
                                            materials.
Types of crystal structures                 Simple cubic, FCC, BCC, diamond, etc.
Metals, insulators, and semiconductors      Metals have a partly-full band of electronic states. Insulators and
                                            semiconductors have a full band separated from an empty band by
                                            an energy gap.
Drude model of conductivity                   ne 2 / m      (13.11)

Electronic collisions in metals             Electron-impurity and electron-phonon scattering
Ohm’s Law                                   R = V/I = constant,  = E/j = constant
The drift velocity                          Average velocity of electrons in a current-carrying wire – very
                                            slow, less than 0.01 mm/s
The Fermi energy                            Energy of conduction electrons in highest filled energy states.

                                            E F  h 2 / 8mr 2   a few eV (13.16)
The Fermi speed                             Very high intrinsic speed of conduction electrons due to Pauli
                                            exclusion principle. vF  106 m/s.
Degeneracy pressure                        Pressure of degenerate electron gas, due to Pauli exclusion
                                            principle. p  (1/3) n EF (13.25)
White dwarfs, nuetron stars, black holes   Stellar remanants: gravity balanced by electron degeneracy
                                            pressure, neutron degeneracy pressure, or nothing.
Classical vs. quantum gases                When wave packets overlap, quantum effects are important.
                                                                   n q   3mkT / h 2 
                                                                                       3/ 2
                                            Crossover at density                              (13.32)

Bose-Einstein Condensation                 Bosons violate Pauli exclusion principle, with a vengence!




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       New Chapter 13, Solids I: Theory                                                  13.38


PROBLEMS FOR CHAPTER 13


SECTION 13.2 (BONDING OF SOLIDS)

13.1 [ = old 17.1 XX]  From the data in Table 13.1 find the cohesive energies of
diamond and germanium in J/mol and kcal/ mol.
13.2 [ = old 17.2 XX]  From the data in Table 13.1 find the cohesive energies of NaCl
and LiF in J/mol and kcal/mol. (Be careful! How many atoms are contained in a mole?)
13.3 [ = old 17.3 XX]  The force between two electric dipoles (each of zero net charge)
results from the imperfect cancellation of the various attractive and repulsive forces
among the four charges involved. (a) Prove that the force between the two dipoles in Fig.
13.37(a) is repulsive. (b) What is it for the dipoles in Fig. 13.37(b)?
13.4 [ = old 17.4 XX]  (a) Prove that the force between two dipoles with two like
charges closest as in Fig. 13.37(a) is repulsive. (b) What is it if unlike charges are closest
as in Fig. 13.38(b)?
13.5 [ = old 17.5 XX]  The van der Waals force between two atoms arises because the
fluctuating dipole moment of atom 1 induces a dipole moment in atom 2, and this
induced moment is always oriented so as to experience an attractive force from atom 1.
This point is illustrated in Fig. 13.2, for one orientation of the dipoles. (a) Sketch the
electric field lines of dipole 1 in Fig. 13.2 and verify that the induced dipole 2 is oriented
as shown. (b) Choose axes in the usual way (x across and y up the page) and put the
origin at dipole 1. If atom 2 is moved onto the y axis, what is the orientation of its
induced dipole moment, and what is the direction of the force on it? (c) Repeat for the
case that atom 2 is put halfway between the x and y axes.


SECTION 13.3 (CRYSTALS AND NONCRYSTALS)

13.6 [ = old 17.6 XX]  One can think of a simple cubic lattice [Fig. 13.5(b)] as a stack of
many unit cubes [Fig. 13.5(c)]. However, in counting atoms there is a danger of mis-
counting, since adjacent cubes share several atoms. (a) Consider any one atom in the
lattice. How many cubes share this atom? (b) Consider any one cube. How many atoms
does the cube have a share of? (c) Prove that in the lattice as a whole, there is one atom
per unit cube.
13.7 [ = old 17.7 XX]  Consider the Na+ ion at the center of the cube shown in Fig.
13.6(b). (a) In terms of the nearest-neighbor distance r0, find the distance of any one of
the next-nearest-neighbor ions. (b) How many of these ions are there? (c) Are they Na+ or
Cl– ?
13.8 [ = old 17.8 XX]  Consider the Na+ ion at the center of the cube shown in Fig.
13.6(b). (a) In terms of the nearest-neighbor distance r0, find the distance of any one of
the third-nearest-neighbor ions. (b) How many of these ions are there? (c) Are they Na+
or Cl–?



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       New Chapter 13, Solids I: Theory                                               13.39

13.9 [ = old 17.9 XX]  The density of solid NaCl is 2.16 g/cm3. Find the
nearest-neighbor distance. (Hint: The number density of ions is 1 ion per r03 .)

13.10 [ = old 17.10 XX]  The KCl crystal has the same FCC structure as NaCl, and the
nearest-neighbor distance is 0.315 nm. What is the density of KCl? (Hint: The number
density of ions is 1 ion per r03 .)

13.11 [ = old 17.11 XX]  In Example 13.1 [Eq. (13.2)] we estimated the potential
energy of any one ion in solid NaCl. (a) Make a corresponding estimate of the PE of any
one ion in KCl, whose crystal structure is the same as NaCl, but whose nearest-neighbor
separation is 0.31 nm. (b) This estimate ignores the repulsive forces that come into play
when the ions get very close. Thus your answer should be somewhat below the observed
value of –7.2 eV. Use this observed value to find the atomic cohesive energy of KCl in
eV/atom. (The ionization energy of K is 4.3 eV; the electron affinity of Cl is 3.6 eV.)
13.12 [ = old 17.12 XX]  Consider the Na+ ion at the center of the cube in Fig. 13.6(b).
(a) How many fourth-nearest neighbors does it have? (b) Are they Na+ or Cl– ? (c) Where
are they? (d) What is their distance?
13.13 [ = old 17.13 XX]  We saw in Example 13.1 that the PE of any one ion in NaCl
has the form U = – ke2/r0 , where  = 1.75 is called the Madelung constant. The calcula-
tion of this constant in three dimensions is much more trouble than it is worth here, but
the one-dimensional. Madelung constant is fairly easy to calculate, as follows: Consider a
long one-dimensional "crystal" of alternating Na+ and Cl– ions spaced a distance r0 apart.
Prove that the PE of any one ion is U = – ke2/r0 where  = 1.39. [Hint: The series for
ln(l + z) in Appendix B (with z = 1) is useful in this calculation.]
13.14 [ = old 17.14 XX]  Extend the reasoning of Problem 13.6 to prove that the
number of atoms per unit cell of the BCC lattice is 2.
13.15 [ = old 17.15 XX]  Extend the reasoning of Problem 13.6 to prove that the
number of atoms per unit cell of the FCC lattice is 4.
13.16 [ = old 17.16 XX]  Gold has the FCC structure, and its density is 19.3 g/cm3. (a)
Use the result of Problem 13.15 to find the edge length of the unit cube. (b) What is the
nearest-neighbor separation (center to center as usual)?
13.17 [ = old 17.17 XX]  Copper has the FCC structure, and the center-to-center
distance between nearest-neighbor atoms is 0.255 nm. Use this information and the result
of Problem 13.15 to find the density of solid copper.
13.18 [ = old 17.18 XX]  Iron has the BCC structure, and its density is 7.86 g/cm3. Use
the result of Problem 13.14 to find the center-to-center distance between nearest-neighbor
iron atoms.
13.19 [ = old 17.19 XX]  In the CsCl crystal the Cs+ ions and Cl– ions are arranged on
two identical simple cubic lattices. The Cl– lattice is offset from the Cs+ lattice, so that
each Cl– is at the exact center of a cube of Cs+ ions, and vice versa. Thus a unit cell of
CsCl looks just like the BCC cell in Fig. 13.7 (except that the center ion is Cl–, whereas
those at the corners are Cs+, or vice versa). (a) The density of CsCl is 3.97 g/cm3. Use the


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       New Chapter 13, Solids I: Theory                                                 13.40

result of Problem 13.14 to find the edge length of the unit cube of CsCl. (b) What is the
nearest-neighbor distance?
13.20 [ = old 17.20 XX]  (a) Imagine a large box full of identical rigid spheres packed
in a simple cubic lattice and each touching all of its nearest neighbors. What fraction of
the total volume is taken up by the spheres? (This fraction is called the packing fraction.)
Answer the same question for (b) FCC and (c) BCC packing. (d) Which arrangement
packs the most spheres per volume?
13.21 [ = old 17.21 XX]  To estimate the tensile strength of the material in Example
13.2, we had to know the force needed to pull two atoms apart. To get an estimate of this
force consider the semiclassical model of the H2 molecule in Fig. 16.34 (Problem 16.11).
Using the result s = R 0 / 3 from Problem 16.11 and taking R0 = 0.3 nm as a
representative lattice spacing, find the total force of one atom on the other.
13.22 [ = old 17.22 XX]  Problem 13.21 suggests one way to estimate the force needed
to separate two bonded atoms; here is another. The graph in Fig. 16.6 shows the energy of
an Na+- Cl– pair as a function of their separation. Since the slope of this graph is the force
between the ions, the maximum force needed to pull them apart is just the maximum
slope of the graph. Use the graph to show that this force is of order 3  10–9 newton.
13.23 [ = old 17.23 XX]  In Example 13. 1, Eq. (13. 1), we wrote down the first three
terms in the electrostatic PE of an ion in the NaCl lattice. What are the next two terms?
13.24 [ = old 17.24 XX]  In Example 13. 1, Eq. (13. 1), we said that the electrostatic
PE of any one ion in NaCl is Ues = – ke2/r, where  = 1.75. This is the electrostatic
energy of simple point charges, and we should have included a contribution due to the
repulsion of the ions at close range. This repulsive contribution can be approximated by a
term of the form Urep = + a/r n, where a and n are positive constants. Thus the total PE is
                                                              ke 2 
                       U (r )  U es (r )  U rep (r )          n
                                                               r   r
(a) At the equilibrium separation r = r0 , this energy is a minimum. Use this fact to find
the constant a, and hence show that the total PE of an ion at equilibrium is

                       U (r0 )  
                                      ke 2   1  1 
                                                   
                                       r0        n
(b) This result differs from our rough estimate (13.2) only by the second term 1/n. For
NaCl, r0 = 0.281 nm and n = 8. Use these values to find the total PE of an ion in NaCl.
Compare with the rough answer of –9.0 eV found in Example 13.1 and the observed
value of – 7.92 eV.


SECTION 13.4 (ENERGY LEVELS OF ELECTRONS IN SOLIDS; BANDS)

13.25 [ = old 17.25 XX]  The density of solid silicon is 2.33 g/cm3. (a) Estimate the total
number, N, of silicon atoms in a microcrystal 1 m  1 m  1 m. (b) The valence band
in silicon is about 1 eV wide. Given that there are 4N states in this band, make a rough


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       New Chapter 13, Solids I: Theory                                                  13.41

estimate of the average separation between states in the valence band of this crystal. (c)
Consider a silicon solar cell that is 1 cm  1 cm.  1 mm; repeat both calculations for
these dimensions.
13.26 [ = old 17.26 XX]  The density of solid germanium is 5.32 g/cm3. (a) Estimate
the total number, N, of atoms in a microcrystal 1 m  1 m  1 m. (b) The conduction
band in germanium is about 2 eV wide. Given that there are 4N states in this band, make
a rough estimate of the average separation between states in the conduction band of this
crystal. (c) Consider a germanium crystal that is 1 cm  1 cm  1 cm; repeat both
calculations for these dimensions.
13.27 [ = old 17.27 XX]  Imagine a crystal containing just one type of atom on a simple
cubic lattice. If the lattice spacing is 0.2 nm and the whole crystal is 3 mm  3 mm  3
mm, how many states are there in the 2p band?
13.28 [ = old 17.28 XX]  Consider a solid containing N atoms and a band that evolved
from an atomic level of angular momentum 1. How many states does this band contain?
Check that your answer gives 2N for an s band and 6N for a p band.


SECTIONS 13.5 AND 13.6 (CONDUCTORS AND INSULATORS)

13.29 [ = old 17.29 XX]  Consider a single crystal of an insulator with a band gap of 3.5
eV. What colors of visible light will this crystal transmit?
13.30 [ = old 17.30 XX]  Crystalline sulfur absorbs blue light but no other colors; as a
result, it is a pale yellow transparent solid. It is also an electrical insulator. From this
information deduce the width of the band gap of crystalline sulfur, and describe the
occupancy of the bands above and below the gap.
13.31 [ = old 17.31 XX]  Germanium is a semiconductor with a band gap of 0.8 eV.
What colors of visible light does it transmit?
13.32 [ = old 17.32 XX]  Explain why many insulators are transparent to visible light,
whereas most semiconductors are opaque to visible light but transparent to infrared.
13.33 [ = old 17.33 XX]  The longest wavelength of electromagnetic radiation that is
absorbed in gallium arsenide is 890 nm. What is the band gap of this semiconductor?
13.34 [ = old 17.38 XX]  Consider a hypothetical element that forms a solid with bands
as shown in Fig. 13.38. (a) Suppose that the isolated atom has configuration1s22s2. If its
equilibrium separation is r0 = a, is the solid a conductor or insulator? What if r0 = b?
Answer both questions for the case that the atomic configuration is (b) 1s22s22p1 and (c)
1s22s22p6.


SECTION 13.6 (DRUDE MODEL OF CONDUCTIVITY)

13.35 [ = new XX]  (a) Make a sketch of current I vs. voltage V for an ohmic resistor.
What is the significance of the slope of your graph? (b) When a 120V light bulb is on,
the temperature of the filament is about 3000 C, and the resistance due to lattice


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       New Chapter 13, Solids I: Theory                                                  13.42

vibrations is approximately 10 times the resistance of the filament at room temperature.
Make a sketch of the current I vs. voltage V for a tungsten light bulb filament over a
range of voltages from 0 to 120 V. (c) Why is a light bulb most likely to burn out,
immediately after it is first turned on?
13.36 [ = new XX]  (a) Consider a simple cubic lattice of atoms, in which the number
density of atoms is na . Show that the distance d between nearest neighbors is given by d
= na -1/3. (b) In a metal, we can define a distance rs as the radius of a sphere whose volume
is the volume per conduction electron. Show that rs is related to the conduction electron
density n by the formula rs   3 / ( 4 n )1/ 3 (c) What is the relation between rs and d for
a monovalent metal with a simple cubic lattice?
13.37 [ = new XX]  Compute the number density n (number per volume) of conduction
electrons in silver, assuming one conduction electron per atom. The mass density of
silver is 10.5 grams/cm3.
13.38 [ = new XX]  Consider a metal in which the distance between nearest-neighbor
atoms is d = 0.3 nm. (a) Assuming a simple cubic lattice structure, and one conduction
electron per atom, what is the conduction electron density n (number per volume)? (b)
Assume that a 1 mm diameter wire made of this metal carries a current of 50 A (a rather
large current). What is the current density j? (c) What is the drift speed in this current-
carrying wire?
13.39 [ = new XX]  In Section 13.8, we show that the mean speed of conduction
electrons in a metal is roughly 0.003 c (this is approximately the Fermi speed). Suppose
that the velocity of a particular conduction electron is known to within 1%, so that the
uncertainty in the speed is approximately v = 3 10–5 c. Use the Heisenberg uncertainty
principle to show that the resulting uncertainty in the position of the electron is greater
than, but not enormously greater than, the distance between atoms in a solid. (Since this
calculation is only approximate, you can assume 1D motion.) This calculation is the
justification for the use of the semi-classical approximation of the Drude model, in which
we assume that quantum mechanical electrons have a well-defined position and
momentum.
13.40 [ = new XX]  Consider a cylindrical resister of length L, cross-sectional area A,
resistivity  , and resistance R. The resistor has a voltage difference V between its ends
and carries a current I. Show that ―Ohm’s Law‖ V = IR is equavalent to E = j , with
resistance R related to the resistivity  by the R =  L/A.
13.41 [ = new XX]  One can define the scattering time  in the following way: the
probablity P that an electron suffers a collision during an infinitesimal time interval dt is
                             P (collision in t ime dt )  dt /  .                     (13.35)
Note that the probability that the electron suffers no collision during a time interval dt is
then given by
                             P (no collision in dt )  1  dt /  .                    (13.36)




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        New Chapter 13, Solids I: Theory                                                          13.43



Consider now the probability P (t ) that a given electron undergoes no collision during a
finite time interval t.
(a) Show that P’ obeys the equation
                                               dP    dt
                                                     .                                        (13.37)
                                                P    
[Hint: First, argue that P (t  dt )  P (t )(1  dt /  ) .]
(b) Using (13.37), solve for P (t ) and sketch this function.
(c) Suppose that a collision occurs at time t = 0. Argue that the probability P (t ) that the
next collision occurs between time t and t + dt is given by
                                                                   dt
                                          P (t ) dt  e t /         .                       (13.38)
                                                                   
(d) Use (13.38), to compute the average time, t , between collisions.

(e) Compute the rms average time             t 2 between collisions.




SECTIONS 13.7 and 13.8 (ELECTRONIC COLLISIONS IN METALS and THE FERMI SPEED)

13.42 [ = new XX]  Fig.13.29 shows the dependence of resistivity on temperature for a
pure metal and a semiconductor. Sketch a new graph, showing the dependence of
conductivity on temperature for a metal and a semiconductor. Be sure to label your
curves clearly.
13.43 [ = new XX]  At high temperatures (T 500 K), the resistivity of tungsten is
approximately proportional to the absolute temperature (  T). A hot tungsten light
bulb filament maintains a stable temperature because the electrical power into the bulb (P
= I 2R = V 2/R) is exactly balanced by the power out due to cooling by radiation and
conduction. At high temperatures, the cooling is dominated by radiation, and the power
radiated is given by the formula , P = A   T 4, where A is the surface area of the body,
and both  and  are constants, called the emissivity and the Stephan-Boltzmann
constant*, respectively. (a) Show that the temperature of a hot light bulb filament with
voltage V is proportional to V 2/5. (b) Show that the temperature of a hot filament carrying
a current I is proportional to I 2/3.
13.44 [ = new XX]  Estimate the conductivity  of an alloy or very dirty metal.
Assume a simple cubic lattice, one conduction electron per atom, and a mean free path of
about 2 lattice constants (twice the distance between atoms). Take the Fermi speed to be


        Be aware: the Greek letter  (sigma) is used to denote both the conductivity and the Stephan-
        *

Boltzmann constant. Don’t confuse them.


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       New Chapter 13, Solids I: Theory                                               13.44

0.003 c, typical of metals. How does this compare to the conductivity of copper at room
temperature?
13.45 [ = new XX]  At room temperature, the mean free path in a clean metal sample is
about 150nm. The ―triple-R‖ of the metal [R(300K)/R(4K) ] is measured to be 100. What
is the mean free path in the metal at T = 4 K?
13.46 [ = new XX]  Repeat the calculation leading to equation (13.16)for the Fermi
energy for the case of a divalent metal (one with two conduction electrons per atom).
(Hint: For a simple cubic lattice, the number density n is related to the nearest-neighbor
separation r by n = 1/r3.)
13.47 [ = new XX]  Estimate the mean free path in a metal with a resistivity of 5
cm (typical of a good metal at room temperature). Assume a typical metal: one
conduction electron per atom, a nearest neighbor atomic separation of 0.3 nm, and a
Fermi speed of 106 m/s.
13.48 [ = new XX]  (a) Using Drude’s formula  = ne2 / m and the known values of
the conductivity of metals, make an order-of-magnitude estimate of the scattering time of
. (b) Compute the thermal speed of an electron at room temperature. Drude assumed,
incorrectly, that the mean speed of conduction electrons is given by their thermal speed.
He also assumed, incorrectly, that the atomic nuclei in the metals scatter conduction
electrons, which implies a mean free path of a few lattice constants. (c) Use the thermal
speed and the scattering time to compute the mean free path. Your answer should be a
distance equal to a few lattice constants, consistent with Drude’s (incorrect) assumption
of atomic scattering. This calculation shows that, although Drude’s made two incorrect
assumptions, his model was self-consistent.


SECTION 13.9 (DEGENERACY PRESSURE )

13.49 [ = new XX]  An particle in a 1D rigid box of length L (an infinite square well) is
in it ground state. The box is compressed to length L/2. (a) By what factor does the
ground-state energy increase? (b) By what factor does the first excited state energy
increase? (c) Repeat the calculations in (a) and (b) for the case of a particle in a 3D box
of volume V = L 3.
13.50 [ = new XX]  Eqn (13.16) is an approximate expression for the Fermi energy in
terms of the mean distance r between nearest neighbor electrons (which is the same as the
distance between nearest-neighbor atoms in a monovalent metal). (a) Rewrite this
expression to give the Fermi energy as a function of the number density of electrons n
(number per volume). (b) Consider a metal with a Fermi energy of 4eV. What is the
Fermi energy if this metal is compressed to twice its normal density? (As the next
problem shows, such compression requires enormous pressure.) Hint: you should be able
to compute the answer without looking up Planck’s constant h or the mass of the election.
13.51 [ = new XX]  (a) An ideal gas at standard temperature and pressure (STP: T =
273K, p = 1 atm = 1.01 × 105 Pa) is compressed to one-half of its original volume while
the temperature is held constant. What is the pressure of the compressed gas? Give the


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       New Chapter 13, Solids I: Theory                                                13.45

answer in atmospheres. (b) Why is the degeneracy pressure unimportant in an ordinary
gas (such as air) at STP? (c) A metal with a Fermi energy of 4eV and an electron
degeneracy pressure of 106 atm is compressed to one-half of its initial volume. What is
the new value of the degeneracy pressure? Hint: use the results of the previous problem.


SECTION 13.10 (WHITE DWARFS, NEUTRON STARS, AND BLACK HOLES )

13.52 [ = new XX]  A neutron star consists of neutrons packed so closely together that
they touch, just as in the interior of an atomic nucleus. Estimate the density of a neutron
star, given that the radius of a neutron is about 1 fm (10-15 m).
13.53 [ = new XX]  Consider a star with a mass 3 times that of the Sun, a rotational
period of 30 days, and an average density equal to that of water (which is a typical density
for a star) . If the star collapses into a neutron star with a diameter of 10 km, what is the
new rotational period? Assume that the original star and the final neutron star both have
uniform densities. Hint: use conservation of angular momentum. (This calculation
produces a higher rotational rate than is usually found in neutron stars. Our calculation is
flawed because normal stars, like the Sun, do not have a uniform density. Instead, most of
the mass is concentrated near the center.)
13.54 [ = new XX]  (a) Use non-relativistic classical mechanics to derive an expression
for the radius of the event horizon surrounding a black hole of mass M. (In this case, a
non-relativistic calculation happens to produce exactly the right answer.) (b) What is the
event horizon radius of a 10 solar mass black hole? (c) What is the acceleration of
gravity at the event horizon of a 10 solar mass black hole?
13.55 [ = new XX]  Derive an expression for the acceleration of gravity at the event
horizon surrounding a black hole of mass M. Curiously, the accelation of gravity at the
event horizon decreases as the mass of the black hole increases. How massive is a black
hole which has an acceleration of 1 g (9.8 m/s2) at its event horizon? Give your answer in
solar masses.
13.56 [ = new XX]  Estimation of the radius of a neutron star. (a) Estimate the
gravitational pressure at the center of a star of mass M and radius R, assuming a uniform
mass density. A fairly accurate estimate can be made from the following rough
calculation: consider the star to be made up of 2 hemispheres each of mass M/2. The
centers of the two hemispheres are roughly a distance R apart. The gravitational force
between the hemispheres is then approximately F = G (M/2)2/R 2. This force is spread
out over the area of contact A between the hemispheres and the pressure at the center is
roughly p = F/A. (b) Starting with equations (13.16) and (13.25), derive an expression for
the neutron degeneracy pressure in terms of R and M. Hint: the mass density  of the star,
the number density n of neutrons, and the neutron mass m are related by  = m n. (c)
The neutron star has a stable size because the inward gravitational pressure is equal to the
outward degeneracy pressure. Derive an expression for the radius of a neutron star of
mass M. If your expression is correct, it should show that the radius R and the mass M
are related by R  M –1/3. A larger mass produces a smaller radius! (d) Calculate the
radius of a 2 solar mass neutron star.


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       New Chapter 13, Solids I: Theory                                               13.46



SECTION 13.12 (BOSE-EINSTEIN CONDENSATION )

13.57 [ = new XX]  Separation of energy levels in a gas. (a) Review Sections( 7.4 and
8.3) and then compute the separation between the ground state energy and the first
excited state energy for a particle in a rigid cubical box of edge length L. (b) Consider a
gas of non-interacting bosons with mass m equal to that of helium atoms, contained in a
cubical box, with edge length L = 1 cm. If no Bose-Einstein condensation were to occur,
how low would the temperature have to be for most of the atoms to be in the ground
state?
13.58 [ = new XX]  Bose-Einstein Condensation in rubidium vapor. Use equation
(13.33) to estimate the temperature at which a gas of 2000 rubidium-87 atoms confined to
a volume of 10-15 m3 will undergo a Bose-Einstein transition. These were the conditions
in the original experiment of Cornell and Wieman.
13.59 [ = new XX]  Critical temperature for superfluid He-4. Liquid 4He undergoes a
transition to a superfluid at a critical temperature Tc , which should be given
approximately by (13.33). [Equation (13.33) was derived for the case of a gas of weakly-
interacting particles and is not expected to be very accurate for a system of strongly-
interacting particles such as liquid helium.] Use (13.33) to get an order-of-magnitude
estimate of the temperature at which liquid helium undergoes a Bose-Einstein transition
and forms a superfluid. The mass density of liquid helium of 0.15 grams/cm3. (The
observed value fo the critical temperature of liquid helium is Tc = 2.17 K.)




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